Section14.6Volume Between Surfaces and Triple Integration
Subsection14.6.1Volume between surfaces
We learned in Section 14.2 how to compute the signed volume \(V\) under a surface \(z=f(x,y)\) over a region \(R\text{:}\)\(V = \iint_R f(x,y)\, dA\text{.}\) It follows naturally that if \(f(x,y)\geq g(x,y)\) on \(R\text{,}\) then the volume between \(f(x,y)\) and \(g(x,y)\) on \(R\) is
\begin{equation*}
V = \iint_R f(x,y)\, dA - \iint_R g(x,y)\, dA = \iint_R \big(f(x,y)-g(x,y)\big)\, dA\text{.}
\end{equation*}
Let \(f\) and \(g\) be continuous functions on a closed, bounded region \(R\text{,}\) where \(f(x,y)\geq g(x,y)\) for all \((x,y)\) in \(R\text{.}\) The volume \(V\) between \(f\) and \(g\) over \(R\) is
\begin{equation*}
V =\iint_R \big(f(x,y)-g(x,y)\big)\, dA\text{.}
\end{equation*}
Find the volume of the space region bounded by the planes \(z=3x+y-4\text{,}\)\(z=8-3x-2y\text{,}\)\(x=0\) and \(y=0\text{.}\) In Figure 14.6.3.(a) the planes are drawn; in Figure 14.6.3.(b), only the defined region is given.
Two planes are shown in space relative to a set of three-dimensional coordinate axes. Each plane is plotted over the same rectangular domain in the \(xy\) plane, and drawn in the shape of a parallelogram.
The planes intersect along a line. With respect to the parallelogram shapes used in the image, one pair of opposite corners for each plane lies along the line of intersection.
This is another drawing of the same two planes as Figure 14.6.3.(a), but this time, the planes are plotted over a triangular domain. The sides of the triangle correspond to the \(x\) and \(y\) axes, and the line of intersection of the two planes.
Each plane is therefore drawn as a triangle in space. The two triangles meet along an edge, and the opposite vertex of each triangle is along the \(z\) axis.
One triangle lies below the other, if we measure height relative to the \(z\) axis. These two triangles make up two of the four faces of a tetrahedron. The other two faces lie in the \(xz\) and \(yz\) coordinate planes, respectively.
We need to determine the region \(R\) over which we will integrate. To do so, we need to determine where the planes intersect. They have common \(z\)-values when \(3x+y-4=8-3x-2y\text{.}\) Applying a little algebra, we have:
The planes intersect along the line \(2x+y=4\text{.}\) Therefore the region \(R\) is bounded by \(x=0\text{,}\)\(y=0\text{,}\) and \(y=4-2x\text{;}\) we can convert these bounds to integration bounds of \(0\leq x\leq 2\text{,}\)\(0\leq y\leq 4-2x\text{.}\) Thus
This no longer looks like a “double integral,” but more like a “triple integral.” Just as our first introduction to double integrals was in the context of finding the area of a plane region, our introduction into triple integrals will be in the context of finding the volume of a space region.
A zoomed in view of a portion of the ellipsoid in Figure 14.6.4.(a) where it meets the positive \(y\) axis. The grid lines on the surface are horizontal (traces with constant \(z\) value) and vertical (traces by planes through the \(z\) axis), like lines of latitude and longitude.
On one of the “rectangles” formed by the grid, a small rectangular box is drawn. The image illustrates how we can approximate the volume of an object like an ellipsoid by dividing it up into small cube-like pieces.
To formally find the volume of a closed, bounded region \(D\) in space, such as the one shown in Figure 14.6.4.(a), we start with an approximation. Break \(D\) into \(n\) rectangular solids; the solids near the boundary of \(D\) may possibly not include portions of \(D\) and/or include extra space. In Figure 14.6.4.(b), we zoom in on a portion of the boundary of \(D\) to show a rectangular solid that contains space not in \(D\text{;}\) as this is an approximation of the volume, this is acceptable and this error will be reduced as we shrink the size of our solids.
The volume \(\Delta V_i\) of the \(i\)th solid \(D_i\) is \(\Delta V_i = \dx_i\dy_i\ddz_i\text{,}\) where \(\dx_i\text{,}\)\(\dy_i\) and \(\ddz_i\) give the dimensions of the rectangular solid in the \(x\text{,}\)\(y\) and \(z\) directions, respectively. By summing up the volumes of all \(n\) solids, we get an approximation of the volume \(V\) of \(D\text{:}\)
\begin{equation*}
V \approx \sum_{i=1}^n \Delta V_i = \sum_{i=1}^n \dx_i\dy_i\ddz_i\text{.}
\end{equation*}
Let \(\norm{\Delta D}\) represent the length of the longest diagonal of rectangular solids in the subdivision of \(D\text{.}\) As \(\norm{\Delta D}\to 0\text{,}\) the volume of each solid goes to 0, as do each of \(\dx_i\text{,}\)\(\dy_i\) and \(\ddz_i\text{,}\) for all \(i\text{.}\) Our calculus experience tells us that taking a limit as \(\norm{\Delta D}\to 0\) turns our approximation of \(V\) into an exact calculation of \(V\text{.}\) Before we state this result in a theorem, we use a definition to define some terms.
Let \(D\) be a closed, bounded region in space. Let \(a\) and \(b\) be real numbers, let \(g_1(x)\) and \(g_2(x)\) be continuous functions of \(x\text{,}\) and let \(f_1(x,y)\) and \(f_2(x,y)\) be continuous functions of \(x\) and \(y\text{.}\)
Our informal understanding of the notation \(\iiint_D\, dV\) is “sum up lots of little volumes over \(D\text{,}\)” analogous to our understanding of \(\iint_R\, dA\) and \(\iint_R\, dm\text{.}\)
Let \(D\) be a closed, bounded region in space and let \(\Delta D\) be any subdivision of \(D\) into \(n\) rectangular solids, where the \(i\)th subregion \(D_i\) has dimensions \(\dx_i\times\dy_i\times\ddz_i\) and volume \(\Delta V_i\text{.}\)
If \(D\) is defined as the region bounded by the planes \(x=a\) and \(x=b\text{,}\) the cylinders \(y=g_1(x)\) and \(y=g_2(x)\text{,}\) and the surfaces \(z=f_1(x,y)\) and \(z=f_2(x,y)\text{,}\) where \(a\lt b\text{,}\)\(g_1(x)\leq g_2(x)\) and \(f_1(x,y)\leq f_2(x,y)\) on \(D\text{,}\) then
\(V\) can be determined using iterated integration with other orders of integration (there are 6 total), as long as \(D\) is defined by the region enclosed by a pair of planes, a pair of cylinders, and a pair of surfaces.
We evaluated the area of a plane region \(R\) by iterated integration, where the bounds were “from curve to curve, then from point to point.” Theorem 14.6.7 allows us to find the volume of a space region with an iterated integral with bounds “from surface to surface, then from curve to curve, then from point to point.” In the iterated integral
the bounds \(a\leq x\leq b\) and \(g_1(x)\leq y\leq g_2(x)\) define a region \(R\) in the \(xy\)-plane over which the region \(D\) exists in space. However, these bounds are also defining surfaces in space; \(x=a\) is a plane and \(y=g_1(x)\) is a cylinder. The combination of these 6 surfaces enclose, and define, \(D\text{.}\)
Example14.6.9.Finding the volume of a space region with triple integration.
Find the volume of the space region in the first octant bounded by the plane \(z=2-y/3-2x/3\text{,}\) shown in Figure 14.6.10, using the order of integration \(dz\, dy\, dx\text{.}\) Set up the triple integrals that give the volume in the other 5 orders of integration.
A plane in space is illustrated, by showing the portion of the plane the lies in the first octant. This portion of the plane is a triangle, whose vertices lie on the coordinate axes. Using the image (or the equation \(z=2-y/3-2x/3\)), we can determine that the plane meets the coordinate axes at the points \((3,0,0)\text{,}\)\((0,6,0)\text{,}\) and \((0,0,2)\text{.}\)
The region in space whose volume is computed in this problem is a tetrahedron, whose faces are the plane described above, as well as the triangles formed in the three coordinate planes by the coordinate axes and the edges of the first triangle.
Starting with the order of integration \(dz\, dy\, dx\text{,}\) we need to first find bounds on \(z\text{.}\) The region \(D\) is bounded below by the plane \(z=0\) (because we are restricted to the first octant) and above by \(z=2-y/3-2x/3\text{;}\)\(0\leq z\leq 2-y/3-2x/3\text{.}\)
To find the bounds on \(y\) and \(x\text{,}\) we “collapse” the region onto the \(xy\)-plane, giving the triangle shown in Figure 14.6.11. (We know the equation of the line \(y=6-2x\) in two ways. First, by setting \(z=0\text{,}\) we have \(0 = 2-y/3-2x/3 \Rightarrow y=6-2x\text{.}\) Secondly, we know this is going to be a straight line between the points \((3,0)\) and \((0,6)\) in the \(xy\)-plane.)
A set of three-dimensional coordinate axes is shown, with the \(z\) axis pointing upward. In the \(xy\) plane a triangle is drawn. The sides of the triangle are formed by the \(x\) axis, the \(y\) axis, and the line from \((3,0,0)\) to \((0,6,0)\text{,}\) which is one edge of the triangle in space shown in Figure 14.6.10. This line is labeled with the equation \(y-6-2x\text{.}\)
This triangle represents the “shadow” of the plane in space on the \(xy\) plane, when viewed directly from above. It will be used to determine the bounds in the triple integral when we integrate first with respect to \(z\text{.}\)
We define that region \(R\text{,}\) in the integration order of \(dy\, dx\text{,}\) with bounds \(0\leq y\leq 6-2x\) and \(0\leq x\leq 3\text{.}\) Thus the volume \(V\) of the region \(D\) is:
Now consider the volume using the order of integration \(dz\, dx\, dy\text{.}\) The bounds on \(z\) are the same as before, \(0\leq z\leq 2-y/3-2x/3\text{.}\) Collapsing the space region on the \(xy\)-plane as shown in Figure 14.6.11, we now describe this triangle with the order of integration \(dx\, dy\text{.}\) This gives bounds \(0\leq x\leq 3-y/2\) and \(0\leq y\leq 6\text{.}\) Thus the volume is given by the triple integral
\begin{equation*}
V = \int_0^6\int_0^{3-\frac12y}\int_0^{2-\frac13y-\frac23x}\, dz\, dx\, dy\text{.}
\end{equation*}
Following our “surface to surface\(\ldots\)” strategy, we need to determine the \(x\)-surfaces that bound our space region. To do so, approach the region “from behind,” in the direction of increasing \(x\text{.}\) The first surface we hit as we enter the region is the \(yz\)-plane, defined by \(x=0\text{.}\) We come out of the region at the plane \(z=2-y/3-2x/3\text{;}\) solving for \(x\text{,}\) we have \(x= 3-y/2-3z/2\text{.}\) Thus the bounds on \(x\) are: \(0\leq x\leq 3-y/2-3z/2\text{.}\)
Now collapse the space region onto the \(yz\)-plane, as shown in Figure 14.6.12.(a). (Again, we find the equation of the line \(z=2-y/3\) by setting \(x=0\) in the equation \(x=3-y/2-3z/2\text{.}\)) We need to find bounds on this region with the order \(dy\, dz\text{.}\) The curves that bound \(y\) are \(y=0\) and \(y=6-3z\text{;}\) the points that bound \(z\) are 0 and 2. Thus the triple integral giving volume is:
A three-dimensional coordinate system is shown, with the \(z\) axis pointing upward. A triangle is plotted in the \(yz\) plane. The edges of the triangle are the \(y\) and \(z\) coordinate axes, and the line from \((0,6,0)\) to \((0,0,2)\text{,}\) where the plane in space from Figure 14.6.10 meets the \(yz\) plane. This line is labeled with the equation \(z=2-y/3\text{.}\)
This is the “shadow” of the region of integration for this problem on the \(yz\) plane, when viewed along the \(x\) axis. It is used to set up the bounds of integration when we integrate first with respect to \(x\text{.}\)
A three-dimensional coordinate system is shown, with the \(z\) axis pointing upward. A triangle is plotted in the \(xz\) plane. The edges of the triangle are the \(x\) and \(z\) coordinate axes, and the line from \((3,0,0)\) to \((0,0,2)\text{,}\) where the plane in space from Figure 14.6.10 meets the \(xz\) plane. This line is labeled with the equation \(z=2-2x/3\text{.}\)
This is the “shadow” of the region of integration for this problem on the \(xz\) plane, when viewed along the \(y\) axis. It is used to set up the bounds of integration when we integrate first with respect to \(y\text{.}\)
The \(x\)-bounds are the same as the order above. We now consider the triangle in Figure 14.6.12.(a) and describe it with the order \(dz\, dy\text{:}\)\(0\leq z\leq 2-y/3\) and \(0\leq y\leq 6\text{.}\) Thus the volume is given by:
We now need to determine the \(y\)-surfaces that determine our region. Approaching the space region from “behind” and moving in the direction of increasing \(y\text{,}\) we first enter the region at \(y=0\text{,}\) and exit along the plane \(z= 2-y/3-2x/3\text{.}\) Solving for \(y\text{,}\) this plane has equation \(y = 6-2x-3z\text{.}\) Thus \(y\) has bounds \(0\leq y\leq 6-2x-3z\text{.}\)
Now collapse the region onto the \(xz\)-plane, as shown in Figure 14.6.12.(b). The curves bounding this triangle are \(z=0\) and \(z=2-2x/3\text{;}\)\(x\) is bounded by the points \(x=0\) to \(x=3\text{.}\) Thus the triple integral giving volume is:
The \(y\)-bounds are the same as in the order above. We now determine the bounds of the triangle in Figure 14.6.12.(b) using the order \(dy\, dx\, dz\text{.}\)\(x\) is bounded by \(x=0\) and \(x=3-3z/2\text{;}\)\(z\) is bounded between \(z=0\) and \(z=2\text{.}\) This leads to the triple integral:
This problem was long, but hopefully useful, demonstrating how to determine bounds with every order of integration to describe the region \(D\text{.}\) In practice, we only need 1, but being able to do them all gives us flexibility to choose the order that suits us best.
In the previous example, we collapsed the surface into the \(x\)-\(y\text{,}\)\(x\)-\(z\text{,}\) and \(yz\)-planes as we determined the “curve to curve, point to point” bounds of integration. Since the surface was a triangular portion of a plane, this collapsing, or projecting, was simple: the projection of a straight line in space onto a coordinate plane is a line.
Example14.6.13.Finding the projection of a curve in space onto the coordinate planes.
Consider the surfaces \(z=3-x^2-y^2\) and \(z=2y\text{,}\) as shown in Figure 14.6.14.(a). The curve of their intersection is shown, along with the projection of this curve into the coordinate planes, shown dashed. Find the equations of the projections into the coordinate planes.
Two surfaces are plotted relative to a set of three-dimensional coordinate axes. One surface is a portion of a downward-opening circular paraboloid; the other is a plane through the origin.
The curve of intersection of the two surfaces in Figure 14.6.14 is plotted relative to a set of three dimensional coordinate axes, with the surfaces removed to highlight the curve.
The two surfaces are \(z=3-x^2-y^2\) and \(z=2y\text{.}\) To find where they intersect, it is natural to set them equal to each other: \(3-x^2-y^2=2y\text{.}\) This is an implicit function of \(x\) and \(y\) that gives all points \((x,y)\) in the \(xy\)-plane where the \(z\) values of the two surfaces are equal.
To project onto the \(xz\)-plane, we do a similar procedure: find the \(x\) and \(z\) values where the \(y\) values on the surface are the same. We start by solving the equation of each surface for \(y\text{.}\) In this particular case, it works well to actually solve for \(y^2\text{:}\)
Finally, to project the curve of intersection into the \(yz\)-plane, we solve equation for \(x\text{.}\) Since \(z=2y\) is a cylinder that lacks the variable \(x\text{,}\) it becomes our equation of the projection in the \(yz\)-plane.
Example14.6.16.Finding the volume of a space region with triple integration.
Set up the triple integrals that find the volume of the space region \(D\) bounded by the surfaces \(x^2+y^2=1\text{,}\)\(z=0\) and \(z=-y\text{,}\) as shown in Figure 14.6.17.(a), with the orders of integration \(dz\, dy\, dx\text{,}\)\(dy\, dx\, dz\) and \(dx\, dz\, dy\text{.}\)
The plane \(z=-y\) meets the \(xy\) plane along the \(x\) axis. The region that lies between these planes, and within the cylinder \(x^2+y^2=1\text{,}\) is a semi-circular wedge. It resembles the shape of a piece cut from a tree that is being cut down.
In three dimensions, the projection of the solid in Figure 14.6.17.(a) onto the \(xy\) plane is shown. The projection forms a semi-circular region in the plane, between the \(x\) axis and the half of the circle \(x^2+y^2=1\) with \(y\leq 0\text{.}\)
The region \(D\) is bounded below by the plane \(z=0\) and above by the plane \(z=-y\text{.}\) The cylinder \(x^2+y^2=1\) does not offer any bounds in the \(z\)-direction, as that surface is parallel to the \(z\)-axis. Thus \(0\leq z\leq -y\text{.}\)
Collapsing the region into the \(xy\)-plane, we get part of the circle with equation \(x^2+y^2=1\) as shown in Figure 14.6.17.(b). As a function of \(x\text{,}\) this half circle has equation \(y=-\sqrt{1-x^2}\text{.}\) Thus \(y\) is bounded below by \(-\sqrt{1-x^2}\) and above by \(y=0\text{:}\)\(-\sqrt{1-x^2}\leq y\leq 0\text{.}\) The \(x\) bounds of the half circle are \(-1\leq x\leq 1\text{.}\) All together, the bounds of integration and triple integral are as follows:
The region is bounded “below” in the \(y\)-direction by the surface \(x^2+y^2=1 \Rightarrow y=-\sqrt{1-x^2}\) and “above” by the surface \(y=-z\text{.}\) Thus the \(y\) bounds are \(-\sqrt{1-x^2}\leq y\leq -z\text{.}\)
Collapsing the region onto the \(xz\)-plane gives the region shown in Figure 14.6.18.(a); this half disk is bounded by \(z=0\) and \(x^2+z^2=1\text{.}\) (We find this curve by solving each surface for \(y^2\text{,}\) then setting them equal to each other. We have \(y^2=1-x^2\) and \(y=-z\Rightarrow y^2=z^2\text{.}\) Thus \(x^2+z^2=1\text{.}\)) It is bounded below by \(x=-\sqrt{1-z^2}\) and above by \(x=\sqrt{1-z^2}\text{,}\) where \(z\) is bounded by \(0\leq z\leq 1\text{.}\) All together, we have:
On a set of three-dimensional coordinate axes, the projection of the solid in Figure 14.6.17.(a) onto the \(xz\) plane is shown. It is a semi-circular region, bounded below by the \(x\) axis, and bounded above by the semi-circle \(x^2+z^2=1\text{,}\)\(z\geq 0\text{.}\)
Using the same three-dimensional coordinate system, the projection of the solid in Figure 14.6.17.(a) onto the \(yz\) plane is shown. This projection is a triangle. The base of the triangle lies along the negative \(y\) axis, from the origin to \((0,-1,0)\text{.}\) Another side is the vertical line \(y=-1\) in the \(yz\) plane, from \((0,-1,0)\) to \((0,-1,1)\text{.}\) The remaining side is the line \(z=-y\text{,}\) from \((-1,0,1)\) to the origin.
\(D\) is bounded below by the surface \(x=-\sqrt{1-y^2}\) and above by \(\sqrt{1-y^2}\text{.}\) We then collapse the region onto the \(yz\)-plane and get the triangle shown in Figure 14.6.18.(b). (The hypotenuse is the line \(z=-y\text{,}\) just as the plane.) Thus \(z\) is bounded by \(0\leq z\leq -y\) and \(y\) is bounded by \(-1\leq y\leq 0\text{.}\) This gives:
The following theorem states two things that should make “common sense” to us. First, using the triple integral to find volume of a region \(D\) should always return a positive number; we are computing volume here, not signed volume. Secondly, to compute the volume of a “complicated” region, we could break it up into subregions and compute the volumes of each subregion separately, summing them later to find the total volume.
Example14.6.20.Finding the volume of a space region with triple integration.
Find the volume of the space region \(D\) bounded by the coordinate planes, \(z=1-x/2\) and \(z=1-y/4\text{,}\) as shown in Figure 14.6.21.(a). Set up the triple integrals that find the volume of \(D\) in all 6 orders of integration.
The region of integration for Example 14.6.20 is a pyramid with a rectangular base. The base lies in the \(xy\) plane, and the peak is on the \(z\) axis at \((0,0,1)\text{.}\) The other four sides are triangles that lie above the \(xy\) plane:
One side is in the \(xz\) plane, with vertices \((0,0,0)\text{,}\)\((2,0,0)\) and \((0,0,1)\text{.}\)
Another side lies in a plane labeled with the equation \(z=1-\frac12 x\text{.}\) The vertices of this triangle are \((2,0,0)\text{,}\)\((2,4,0)\text{,}\) and \((0,0,1)\text{.}\)
The last side lies in a plane labeled with the equation \(z=1-\frac14 y\text{.}\) The vertices of this triangle are \((0,4,0)\text{,}\)\((2,4,0)\text{,}\) and \((0,0,1)\text{.}\)
A dashed line in the plane runs from the origin to the opposite corner of the rectangle, at the point \((2,4,0)\text{.}\) This line divides the rectangle into two triangles. One triangle lies between the \(x\) axis and the dashed line, and is labeled \(R_1\text{.}\) The other triangle lies between the \(y\) axis and the dashed line, and is labeled \(R_2\text{.}\)
Following the bounds-determining strategy of “surface to surface, curve to curve, and point to point,” we can see that the most difficult orders of integration are the two in which we integrate with respect to \(z\) first, for there are two “upper” surfaces that bound \(D\) in the \(z\)-direction. So we start by noting that we have
We now collapse the region \(D\) onto the \(xy\)-axis, as shown in Figure 14.6.21.(b). The boundary of \(D\text{,}\) the line from \((0,0,1)\) to \((2,4,0)\text{,}\) is shown in Figure 14.6.21.(b) as a dashed line; it has equation \(y=2x\text{.}\) (We can recognize this in two ways: one, in collapsing the line from \((0,0,1)\) to \((2,4,0)\) onto the \(xy\)-plane, we simply ignore the \(z\)-values, meaning the line now goes from \((0,0)\) to \((2,4)\text{.}\) Secondly, the two surfaces meet where \(z=1-x/2\) is equal to \(z=1-y/4\text{:}\) thus \(1-x/2=1-y/4 \Rightarrow y=2x\text{.}\))
All that is left is to determine bounds of \(R_1\) and \(R_2\text{,}\) depending on whether we are integrating with order \(dx\, dy\) or \(dy\, dx\text{.}\) We give the final integrals here, leaving it to the reader to confirm these results.
The remaining four orders of integration do not require a sum of triple integrals. In Figure 14.6.22 we show \(D\) collapsed onto the other two coordinate planes. Using these graphs, we give the final orders of integration here, again leaving it to the reader to confirm these results.
In a three-dimensional coordinate system, the side of the pyramid in Figure 14.6.21.(a) in the \(xz\) plane is shown. This represents the projection of the solid onto the \(xz\) plane, and it is a triangle, with vertices \((0,0,0)\text{,}\)\((2,0,0)\text{,}\) and \((0,0,1)\text{.}\)
In a three-dimensional coordinate system, the side of the pyramid in Figure 14.6.21.(a) in the \(yz\) plane is shown. This represents the projection of the solid onto the \(yz\) plane, and it is a triangle, with vertices \((0,0,0)\text{,}\)\((0,4,0)\text{,}\) and \((0,0,1)\text{.}\)
Example14.6.23.Finding the volume of a space region.
Set up a triple integral that gives the volume of the space region \(D\) bounded by \(z= 2x^2+2\) and \(z=6-2x^2-y^2\text{.}\) These surfaces are plotted in Figure 14.6.24.(a) and Figure 14.6.24.(b), respectively; the region \(D\) is shown in Figure 14.6.24.(c).
A three-dimensional sketch of a surface in space. The surface is the elliptic paraboloid \(z=6-2x^2-y^2\text{,}\) which opens downward and has its vertex at \((0,0,6)\text{.}\)
The surface \(z=2x^2+2\) is a parabolic cylinder: it has the shape of a long trough with parabolic cross-sections. It opens upward, and is symmetric about the \(y\) axis.
The overall shape of the solid is interesting. The portion of the parabolic cylinder on the bottom looks like a taco shell. The paraboloid on top is like an elongated dome. Perhaps the shape is what you get if you completely fill a taco with a scoop of ice cream, or make a really thick madeline cookie.
The main point of this example is this: integrating with respect to \(z\) first is rather straightforward; integrating with respect to \(x\) first is not.
The bounds on \(z\) are clearly \(2x^2+2\leq z\leq 6-2x^2-y^2\text{.}\) Collapsing \(D\) onto the \(xy\)-plane gives the ellipse shown in Figure 14.6.24.(c). The equation of this ellipse is found by setting the two surfaces equal to each other:
Integrating with respect to \(y\) is not too difficult. Since the surface \(z=2x^2+2\) is a cylinder whose directrix is the \(y\)-axis, it does not create a border for \(y\text{.}\) The paraboloid \(z=6-2x^2-y^2\) does; solving for \(y\text{,}\) we get the bounds
Collapsing \(D\) onto the \(xz\)-plane gives the region shown in Figure 14.6.25.(a); the lower curve is from the cylinder, with equation \(z=2x^2+2\text{.}\) The upper curve is from the paraboloid; with \(y=0\text{,}\) the curve is \(z=6-2x^2\text{.}\) Thus bounds on \(z\) are \(2x^2+2\leq z\leq 6-2x^2\text{;}\) the bounds on \(x\) are \(-1\leq x\leq 1\text{.}\) Thus we have:
In a three-dimensional coordinate system, a sketch is given of the shadow obtained by projecting the solid in Figure 14.6.24.(c) onto the \(xz\) plane.
It is a plane region, drawn in perspective. The region is bounded above by a parabola opening downward, with its vertex at \((0,0,6)\text{,}\) and is bounded below by a parabola opening upward, with its vertex at \((0,0,2)\text{.}\)
In a three-dimensional coordinate system, a sketch is given of the shadow obtained by projecting the solid in Figure 14.6.24.(c) onto the \(yz\) plane.
It is a plane region, drawn in perspective. The region is divided into two parts. The bottom part is labeled \(R_1\text{.}\) It is bounded below by a horizontal line parallel to the \(y\) axis, with \(z=2\text{.}\) This corresponds to the bottom of the parabolic cylinder. The upper bound of \(R_1\) is a parabola that opens downward, with its vertex at \((0,0,4)\text{.}\) This parabola comes from the curve that is formed by the intersection of the two surfaces; it is labeled with the equation \(z=4-y^2/2\text{.}\)
The upper part is labeled \(R_2\text{.}\) This region is bounded below by the parabola that forms the upper bound of \(R_1\text{.}\) It is bounded above by a second parabola that also opens downward, with its vertex at \((0,0,6)\text{.}\) This parabola corresponds to the \(x=0\) trace through the elliptic paraboloid. It is labeled with the equation \(z=6-y^2\text{.}\)
This order takes more effort as \(D\) must be split into two subregions. The two surfaces create two sets of upper/lower bounds in terms of \(x\text{;}\) the cylinder creates bounds
Collapsing \(D\) onto the \(yz\)-axes gives the regions shown in Figure 14.6.25.(b). We find the equation of the curve \(z=4-y^2/2\) by noting that the equation of the ellipse seen in Figure 14.6.24.(c) has equation
\begin{equation*}
x^2+y^2/4=1 \Rightarrow x = \sqrt{1-y^2/4}\text{.}
\end{equation*}
If all one wanted to do in Example 14.6.23 was find the volume of the region \(D\text{,}\) one would have likely stopped at the first integration setup (with order \(dz\, dy\, dx\)) and computed the volume from there. However, we included the other two methods 1) to show that it could be done, “messy” or not, and 2) because sometimes we “have” to use a less desirable order of integration in order to actually integrate.
Subsection14.6.2Triple Integration and Functions of Three Variables
There are uses for triple integration beyond merely finding volume, just as there are uses for integration beyond “area under the curve.” These uses start with understanding how to integrate functions of three variables, which is effectively no different than integrating functions of two variables. This leads us to a definition, followed by an example.
Let \(D\) be a closed, bounded region in space, over which \(g_1(x)\text{,}\)\(g_2(x)\text{,}\)\(f_1(x,y)\text{,}\)\(f_2(x,y)\) and \(h(x,y,z)\) are all continuous, and let \(a\) and \(b\) be real numbers.
We now know how to evaluate a triple integral of a function of three variables; we do not yet understand what it means. We build up this understanding in a way very similar to how we have understood integration and double integration.
Let \(h(x,y,z)\) be a continuous function of three variables, defined over some space region \(D\text{.}\) We can partition \(D\) into \(n\) rectangular-solid subregions, each with dimensions \(\dx_i\times\dy_i\times\ddz_i\text{.}\) Let \((x_i,y_i,z_i)\) be some point in the \(i\)th subregion, and consider the product \(h(x_i,y_i,z_i)\dx_i\dy_i\ddz_i\text{.}\) It is the product of a function value (that’s the \(h(x_i,y_i,z_i)\) part) and a small volume \(\Delta V_i\) (that’s the \(\dx_i\dy_i\ddz_i\) part). One of the simplest understanding of this type of product is when \(h\) describes the density of an object, for then \(h\times\text{ volume } =\text{ mass }\text{.}\)
We can sum up all \(n\) products over \(D\text{.}\) Again letting \(\norm{\Delta D}\) represent the length of the longest diagonal of the \(n\) rectangular solids in the partition, we can take the limit of the sums of products as \(\norm{\Delta D}\to 0\text{.}\) That is, we can find
While this limit has lots of interpretations depending on the function \(h\text{,}\) in the case where \(h\) describes density, \(S\) is the total mass of the object described by the region \(D\text{.}\)
We now use the above limit to define the triple integral, give a theorem that relates triple integrals to iterated iteration, followed by the application of triple integrals to find the centers of mass of solid objects.
Let \(w=h(x,y,z)\) be a continuous function over a closed, bounded region \(D\) in space, and let \(\Delta D\) be any partition of \(D\) into \(n\) rectangular solids with volume \(\Delta V_i\text{.}\) The triple integral of \(h\) over \(D\) is
\begin{equation*}
\iiint_Dh(x,y,z)\, dV = \lim_{\norm{\Delta D}\to 0}\sum_{i=1}^n h(x_i,y_i,z_i)\Delta V_i\text{.}
\end{equation*}
Let \(w=h(x,y,z)\) be a continuous function over a closed, bounded region \(D\) in space, and let \(\Delta D\) be any partition of \(D\) into \(n\) rectangular solids with volume \(V_i\text{.}\)
The limit \(\lim\limits_{\norm{\Delta D}\to 0}\sum_{i=1}^n h(x_i,y_i,z_i)\Delta V_i\) exists.
If \(D\) is defined as the region bounded by the planes \(x=a\) and \(x=b\text{,}\) the cylinders \(y=g_1(x)\) and \(y=g_2(x)\text{,}\) and the surfaces \(z=f_1(x,y)\) and \(z=f_2(x,y)\text{,}\) where \(a\lt b\text{,}\)\(g_1(x)\leq g_2(x)\) and \(f_1(x,y)\leq f_2(x,y)\) on \(D\text{,}\) then
Note: In an aside in Section 14.2, we showed how the summation of rectangles over a region \(R\) in the plane could be viewed as a double sum, leading to the double integral. Likewise, we can view the sum
Here we fix a \(k\) value, which establishes the \(z\)-height of the rectangular solids on one “level” of all the rectangular solids in the space region \(D\text{.}\) The inner double summation adds up all the volumes of the rectangular solids on this level, while the outer summation adds up the volumes of each level.
This triple summation understanding leads to the \(\iiint_D\) notation of the triple integral, as well as the method of evaluation shown in Theorem 14.6.29.
Example14.6.31.Finding the center of mass of a solid.
Find the mass and center of mass of the solid represented by the space region bounded by the coordinate planes and \(z=2-y/3-2x/3\text{,}\) shown in Figure 14.6.32, with constant density \(\delta(x,y,z)=3\,\text{g/cm}^3\text{.}\) (Note: this space region was used in Example 14.6.9.)
A plane in space is illustrated, by showing the portion of the plane the lies in the first octant. This portion of the plane is a triangle, whose vertices lie on the coordinate axes. Using the image (or the equation \(z=2-y/3-2x/3\)), we can determine that the plane meets the coordinate axes at the points \((3,0,0)\text{,}\)\((0,6,0)\text{,}\) and \((0,0,2)\text{.}\)
The region in space whose volume is computed in this problem is a tetrahedron, whose faces are the plane described above, as well as the triangles formed in the three coordinate planes by the coordinate axes and the edges of the first triangle.
We apply Definition 14.6.30. In Example 14.6.9, we found bounds for the order of integration \(dz\, dy\, dx\) to be \(0\leq z\leq 2-y/3-2x/3\text{,}\)\(0\leq y\leq 6-2x\) and \(0\leq x\leq 3\text{.}\) We find the mass of the object:
The evaluation of the triple integral is done in Example 14.6.9, so we skipped those steps above. Note how the mass of an object with constant density is simply “density×volume.”
Example14.6.33.Finding the center of mass of a solid.
Find the center of mass of the solid represented by the region bounded by the planes \(z=0\) and \(z=-y\) and the cylinder \(x^2+y^2=1\text{,}\) shown in Figure 14.6.34, with density function \(\delta(x,y,z) = 10+x^2+5y-5z\text{.}\) (Note: this space region was used in Example 14.6.16.)
The plane \(z=-y\) meets the \(xy\) plane along the \(x\) axis. The region that lies between these planes, and within the cylinder \(x^2+y^2=1\text{,}\) is a semi-circular wedge. It resembles the shape of a piece cut from a tree that is being cut down.
As we start, consider the density function. It is symmetric about the \(yz\)-plane, and the farther one moves from this plane, the denser the object is. The symmetry indicates that \(\overline x\) should be 0.
Though none of the integrals needed to compute the center of mass are particularly hard, they do require a number of steps. We emphasize here the importance of knowing how to set up the proper integrals; in complex situations we can appeal to technology for a good approximation, if not the exact answer. We use the order of integration \(dz\, dy\, dx\text{,}\) using the bounds found in Example 14.6.16. (As these are the same for all four triple integrals, we explicitly show the bounds only for \(M\text{.}\))
As stated before, there are many uses for triple integration beyond finding volume. When \(h(x,y,z)\) describes a rate of change function over some space region \(D\text{,}\) then \(\ds \iiint_D h(x,y,z)\, dV\) gives the total change over \(D\text{.}\) Our one specific example of this was computing mass; a density function is simply a “rate of mass change per volume” function. Integrating density gives total mass.
While knowing how to integrate is important, it is arguably much more important to know how to set up integrals. It takes skill to create a formula that describes a desired quantity; modern technology is very useful in evaluating these formulas quickly and accurately.
In Section 14.7, we learn about two new coordinate systems (each related to polar coordinates) that allow us to integrate over closed regions in space more easily than when using rectangular coordinates.
Two functions \(f_1(x,y)\) and \(f_2(x,y)\) and a region \(R\) in the \(x\text{,}\)\(y\) plane are given. Set up and evaluate the double integral that finds the volume between the surfaces given by the graphs of these two functions over \(R\text{.}\)
In the following exercises, a domain \(D\) is described by its bounding surfaces, along with a graph. Set up the triple integrals that give the volume of \(D\) in all 6 orders of integration, and find the volume of \(D\) by evaluating the indicated triple integral.
The solid is a tetrahedron, plotted in the first octant relative to the usual three-dimensional coordinate axes. The vertices of the tetrahedron are at \((0,0,0)\text{,}\)\((3,0,0)\text{,}\)\((0,1,0)\text{,}\) and \((0,0,2)\text{.}\) Three of the four faces lie in the coordinate planes; the remaining face is in the first octant, and is labeled with the equation \(z=2-\frac23 x-2y\text{.}\)
The solid is a triangular prism. The base is a square, and it lies in the \(xy\) plane. The corners of the square are at \((1,0,0)\text{,}\)\((3,0,0)\text{,}\)\((3,2,0)\text{,}\) and \((1,2,0)\text{.}\)
The two triangular faces are parallel to the \(xz\) plane. One face is in the plane \(y=0\text{,}\) with vertices \((1,0,0)\text{,}\)\((3,0,0)\text{,}\) and \((3,0,1)\text{.}\) The other face is in the plane \(y=2\text{,}\) with vertices \((1,2,0)\text{,}\)\((3,2,0)\text{,}\) and \((3,2,1)\text{.}\)
The front face (according to how the solid is oriented in the image) is a rectangle in the plane \(z=\frac12(3-x)\text{.}\) Two edges are parallel to the \(y\) axis, and two edges are parallel to the hypotenuse of the triangular faces.
The sides are negatively sloped, relative to the \(yz\text{:}\) the equation of the plane is \(z=-y\text{,}\) and setting \(x=0\) or \(x=2\) in this plane produces the sides.
The other surface is a parabolic cylinder. Cross-sections are the half of the cylinder \(z=\frac12 y^2\) with \(y\leq 0\text{.}\) This surface meets the rectangle in the plane at the top and bottom edges of the rectangle. It has the appearance of a sheet that has been hung below the rectangle and attached at either end.
Projecting the solid onto either the \(xy\) or \(xz\) planes produces a rectangle, while projection onto the \(yz\) plane produces a region that is bounded above by the line \(z=-y\text{,}\) and below by the parabola \(z=\frac12 y^2\text{.}\)
The projection of the solid onto the \(xz\) plane is a quarter ellipse. It is the region bounded by the ellipse \(9x^2+z^2=81\text{,}\) and the positive \(x\) and \(z\) axes.
The image contains the usual three-dimensional coordinate axes, and a tetrahedron. Three of the faces of the tetrahedron are parallel to the coordinate planes.
The bottom is a triangle in the \(xy\) plane with vertices \((2,0,0)\text{,}\)\((2,1,0)\text{,}\) and \((0,1,0)\text{.}\)
A cylindrical wedge, between the planes \(z=0\) and \(z=2y\text{.}\) These planes intersect along the \(x\) axis, forming the sharp edge of the wedge. The round face of the wedge is the parabolic cylinder \(y=4-x^2\text{,}\) for \(y\geq 0\text{.}\)
Another side of the solid lies in the \(yz\) plane, forming a region bounded by the \(y\) and \(z\) coordinate axes and the parabola \(y=1-z^2\text{.}\)
When setting up integrals where the integration is first with respect to \(y\text{,}\) it is important to note that the square in the \(xz\) plane needs to be divided into two triangles along the line \(x=z\text{.}\) For \(z\geq x\text{,}\)\(y\) is bounded above by the cylinder \(y=1-z^2\text{.}\) For \(z\leq x\text{,}\)\(y\) is bounded above by the cylinder \(y=1-x^2\text{.}\)
The solid for this exercise is a pyramid. The base in the \(xy\) plane is a rectangle, with \(x\) between \(0\) and \(1\text{,}\) and \(y\) between \(0\) and \(3\text{.}\)
When integrating first with respect to \(z\text{,}\) note that the rectangle in the \(xy\) plane needs to be divided into two triangles along the line \(y=3x\text{.}\) For \(y\geq 3x\text{,}\) the upper bound is the plane \(z=1-y/3\text{.}\) For \(y\leq 3x\text{,}\) the upper bound is the plane \(z=1-x\text{.}\)
In the following exercises, find the center of mass of the solid represented by the indicated space region \(D\) with density function \(\delta(x,y,z)\text{.}\)
