maths.free › Calculus › Applications of Integration › Volumes of Revolution: Cylindrical Shells
Volumes of Revolution: Cylindrical Shells
Calculate the volume of a solid of revolution by using the method of cylindrical shells.
The Method of Cylindrical Shells
Again, we are working with a solid of revolution. As before, we define a region \(R,\) bounded above by the graph of a function \(y=f(x),\) below by the \(x\text{-axis,}\) and on the left and right by the lines \(x=a\) and \(x=b,\) respectively, as shown in (a). We then revolve this region around the y-axis, as shown in (b). Note that this is different from what we have done before. Previously, regions defined in terms of functions of \(x\) were revolved around the \(x\text{-axis}\) or a line parallel to it.
As we have done many times before, partition the interval \([a,b]\) using a regular partition, \(P=\{{x}_{0},{x}_{1}\text{,}\ldots ,{x}_{n}\}\) and, for \(i=1,2\text{,}\ldots ,n,\) choose a point \({x}_{i}^{*}\in [{x}_{i-1},{x}_{i}].\) Then, construct a rectangle over the interval \([{x}_{i-1},{x}_{i}]\) of height \(f({x}_{i}^{*})\) and width \(\Delta x.\) A representative rectangle is shown in (a). When that rectangle is revolved around the y-axis, instead of a disk or a washer, we get a cylindrical shell, as shown in the following figure.
To calculate the volume of this shell, consider .
The shell is a cylinder, so its volume is the cross-sectional area multiplied by the height of the cylinder. The cross-sections are annuli (ring-shaped regions, essentially, circles with a hole in the center), with outer radius \({x}_{i}\) and inner radius \({x}_{i-1}.\) Thus, the cross-sectional area is \(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}.\) The height of the cylinder is \(f({x}_{i}^{*}).\) Then the volume of the shell is
\[\begin{array}{ll}{V}_{\text{shell}} & =f({x}_{i}^{*})(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}) \\ & =\pi f({x}_{i}^{*})({x}_{i}^{2}-{x}_{i-1}^{2}) \\ & =\pi f({x}_{i}^{*})({x}_{i}+{x}_{i-1})({x}_{i}-{x}_{i-1}) \\ & =2\pi f({x}_{i}^{*})(\frac{{x}_{i}+{x}_{i-1}}{2})({x}_{i}-{x}_{i-1}).\end{array}\]Note that \({x}_{i}-{x}_{i-1}=\Delta x,\) so we have
\[{V}_{\text{shell}}=2\pi f({x}_{i}^{*})(\frac{{x}_{i}+{x}_{i-1}}{2})\Delta x.\]Furthermore, \(\frac{{x}_{i}+{x}_{i-1}}{2}\) is both the midpoint of the interval \([{x}_{i-1},{x}_{i}]\) and the average radius of the shell, and we can approximate this by \({x}_{i}^{*}.\) We then have
\[{V}_{\text{shell}}\approx 2\pi f({x}_{i}^{*}){x}_{i}^{*}\Delta x.\]Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate ().
\[{V}_{\text{shell}}\approx f({x}_{i}^{*})(2\pi {x}_{i}^{*})\Delta x,\]\[V\approx \sum _{i=1}^{n}(2\pi {x}_{i}^{*}f({x}_{i}^{*})\Delta x).\]\[V=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}(2\pi {x}_{i}^{*}f({x}_{i}^{*})\Delta x)={\int }_{a}^{b}(2\pi xf(x))dx.\]\[V={\int }_{a}^{b}(2\pi (x+k)f(x))dx.\]Condensed: the full section is in OpenStax Calculus Volume 2.
Which Method Should We Use?
We have studied several methods for finding the volume of a solid of revolution, but how do we know which method to use? It often comes down to a choice of which integral is easiest to evaluate. describes the different approaches for solids of revolution around the \(x\text{-axis}.\) It’s up to you to develop the analogous table for solids of revolution around the \(y\text{-axis}.\)
Let’s take a look at a couple of additional problems and decide on the best approach to take for solving them.
Example
Try it.
For each of the following problems, select the best method to find the volume of a solid of revolution generated by revolving the given region around the \(x\text{-axis},\) and set up the integral to find the volume (do not evaluate the integral).
- The region bounded by the graphs of \(y=x,\) \(y=2-x,\) and the \(x\text{-axis}.\)
- The region bounded by the graphs of \(y=4x-{x}^{2}\) and the \(x\text{-axis}.\)
Solution
- First, sketch the region and the solid of revolution as shown.
Looking at the region, if we want to integrate with respect to \(x,\) we would have to break the integral into two pieces, because we have different functions bounding the region over \([0,1]\) and \([1,2].\) In this case, using the disk method, we would have
\[V={\int }_{0}^{1}(\pi {x}^{2})dx+{\int }_{1}^{2}(\pi {(2-x)}^{2})dx.\]
If we used the shell method instead, we would use functions of \(y\) to represent the curves, producing
\[\begin{array}{ll}V & ={\int }_{0}^{1}(2\pi y[(2-y)-y])dy \\ & ={\int }_{0}^{1}(2\pi y[2-2y])dy.\end{array}\]
Neither of these integrals is particularly onerous, but since the shell method requires only one integral, and the integrand requires less simplification, we should probably go with the shell method in this case. - First, sketch the region and the solid of revolution as shown.
Looking at the region, it would be problematic to define a horizontal rectangle; the region is bounded on the left and right by the same function. Therefore, we can dismiss the method of shells. The solid has no cavity in the middle, so we can use the method of disks. Then
\[V={\int }_{0}^{4}\pi {(4x-{x}^{2})}^{2}dx.\]
Key Concepts
- The method of cylindrical shells is another method for using a definite integral to calculate the volume of a solid of revolution. This method is sometimes preferable to either the method of disks or the method of washers because we integrate with respect to the other variable. In some cases, one integral is substantially more complicated than the other.
- The geometry of the functions and the difficulty of the integration are the main factors in deciding which integration method to use.
Volumes of Revolution: Cylindrical Shells
For the following exercises, find the volume generated when the region between the two curves is rotated around the given axis. Use both the shell method and the washer method. Use technology to graph the functions and draw a typical slice by hand.
For the following exercises, use shells to find the volumes of the given solids. Note that the rotated regions lie between the curve and the \(x\text{-axis}\) and are rotated around the \(y\text{-axis}.\)
For the following exercises, use shells to find the volume generated by rotating the regions between the given curve and \(y=0\) around the \(x\text{-axis}.\)
For the following exercises, find the volume generated when the region between the curves is rotated around the given axis.
For the following exercises, use technology to graph the region. Determine which method you think would be easiest to use to calculate the volume generated when the function is rotated around the specified axis. Then, use your chosen method to find the volume.
For the following exercises, use the method of shells to approximate the volumes of some common objects, which are pictured in accompanying figures.
The Method of Cylindrical Shells
Again, we are working with a solid of revolution. As before, we define a region \(R,\) bounded above by the graph of a function \(y=f(x),\) below by the \(x\text{-axis,}\) and on the left and right by the lines \(x=a\) and \(x=b,\) respectively, as shown in (a). We then revolve this region around the y-axis, as shown in (b). Note that this is different from what we have done before. Previously, regions defined in terms of functions of \(x\) were revolved around the \(x\text{-axis}\) or a line parallel to it.
As we have done many times before, partition the interval \([a,b]\) using a regular partition, \(P=\{{x}_{0},{x}_{1}\text{,}\ldots ,{x}_{n}\}\) and, for \(i=1,2\text{,}\ldots ,n,\) choose a point \({x}_{i}^{*}\in [{x}_{i-1},{x}_{i}].\) Then, construct a rectangle over the interval \([{x}_{i-1},{x}_{i}]\) of height \(f({x}_{i}^{*})\) and width \(\Delta x.\) A representative rectangle is shown in (a). When that rectangle is revolved around the y-axis, instead of a disk or a washer, we get a cylindrical shell, as shown in the following figure.
To calculate the volume of this shell, consider .
The shell is a cylinder, so its volume is the cross-sectional area multiplied by the height of the cylinder. The cross-sections are annuli (ring-shaped regions, essentially, circles with a hole in the center), with outer radius \({x}_{i}\) and inner radius \({x}_{i-1}.\) Thus, the cross-sectional area is \(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}.\) The height of the cylinder is \(f({x}_{i}^{*}).\) Then the volume of the shell is
\[\begin{array}{ll}{V}_{\text{shell}} & =f({x}_{i}^{*})(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}) \\ & =\pi f({x}_{i}^{*})({x}_{i}^{2}-{x}_{i-1}^{2}) \\ & =\pi f({x}_{i}^{*})({x}_{i}+{x}_{i-1})({x}_{i}-{x}_{i-1}) \\ & =2\pi f({x}_{i}^{*})(\frac{{x}_{i}+{x}_{i-1}}{2})({x}_{i}-{x}_{i-1}).\end{array}\]Note that \({x}_{i}-{x}_{i-1}=\Delta x,\) so we have
\[{V}_{\text{shell}}=2\pi f({x}_{i}^{*})(\frac{{x}_{i}+{x}_{i-1}}{2})\Delta x.\]Furthermore, \(\frac{{x}_{i}+{x}_{i-1}}{2}\) is both the midpoint of the interval \([{x}_{i-1},{x}_{i}]\) and the average radius of the shell, and we can approximate this by \({x}_{i}^{*}.\) We then have
\[{V}_{\text{shell}}\approx 2\pi f({x}_{i}^{*}){x}_{i}^{*}\Delta x.\]Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate ().
\[{V}_{\text{shell}}\approx f({x}_{i}^{*})(2\pi {x}_{i}^{*})\Delta x,\]\[V\approx \sum _{i=1}^{n}(2\pi {x}_{i}^{*}f({x}_{i}^{*})\Delta x).\]\[V=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}(2\pi {x}_{i}^{*}f({x}_{i}^{*})\Delta x)={\int }_{a}^{b}(2\pi xf(x))dx.\]\[V={\int }_{a}^{b}(2\pi (x+k)f(x))dx.\]Condensed: the full section is in OpenStax Calculus Volume 1.
Which Method Should We Use?
We have studied several methods for finding the volume of a solid of revolution, but how do we know which method to use? It often comes down to a choice of which integral is easiest to evaluate. describes the different approaches for solids of revolution around the \(x\text{-axis}.\) It’s up to you to develop the analogous table for solids of revolution around the \(y\text{-axis}.\)
Let’s take a look at a couple of additional problems and decide on the best approach to take for solving them.
Example
Try it.
For each of the following problems, select the best method to find the volume of a solid of revolution generated by revolving the given region around the \(x\text{-axis},\) and set up the integral to find the volume (do not evaluate the integral).
- The region bounded by the graphs of \(y=x,\) \(y=2-x,\) and the \(x\text{-axis}.\)
- The region bounded by the graphs of \(y=4x-{x}^{2}\) and the \(x\text{-axis}.\)
Solution
- First, sketch the region and the solid of revolution as shown.
Looking at the region, if we want to integrate with respect to \(x,\) we would have to break the integral into two pieces, because we have different functions bounding the region over \([0,1]\) and \([1,2].\) In this case, using the disk method, we would have
\[V={\int }_{0}^{1}(\pi {x}^{2})dx+{\int }_{1}^{2}(\pi {(2-x)}^{2})dx.\]
If we used the shell method instead, we would use functions of \(y\) to represent the curves, producing
\[\begin{array}{ll}V & ={\int }_{0}^{1}(2\pi y[(2-y)-y])dy \\ & ={\int }_{0}^{1}(2\pi y[2-2y])dy.\end{array}\]
Neither of these integrals is particularly onerous, but since the shell method requires only one integral, and the integrand requires less simplification, we should probably go with the shell method in this case. - First, sketch the region and the solid of revolution as shown.
Looking at the region, it would be problematic to define a horizontal rectangle; the region is bounded on the left and right by the same function. Therefore, we can dismiss the method of shells. The solid has no cavity in the middle, so we can use the method of disks. Then
\[V={\int }_{0}^{4}\pi {(4x-{x}^{2})}^{2}dx.\]
Key Concepts
- The method of cylindrical shells is another method for using a definite integral to calculate the volume of a solid of revolution. This method is sometimes preferable to either the method of disks or the method of washers because we integrate with respect to the other variable. In some cases, one integral is substantially more complicated than the other.
- The geometry of the functions and the difficulty of the integration are the main factors in deciding which integration method to use.
Volumes of Revolution: Cylindrical Shells
For the following exercises, find the volume generated when the region between the two curves is rotated around the given axis. Use both the shell method and the washer method. Use technology to graph the functions and draw a typical slice by hand.
For the following exercises, use shells to find the volumes of the given solids. Note that the rotated regions lie between the curve and the \(x\text{-axis}\) and are rotated around the \(y\text{-axis}.\)
For the following exercises, use shells to find the volume generated by rotating the regions between the given curve and \(y=0\) around the \(x\text{-axis}.\)
For the following exercises, find the volume generated when the region between the curves is rotated around the given axis.
For the following exercises, use technology to graph the region. Determine which method you think would be easiest to use to calculate the volume generated when the function is rotated around the specified axis. Then, use your chosen method to find the volume.
For the following exercises, use the method of shells to approximate the volumes of some common objects, which are pictured in accompanying figures.
现在轮到你 没有计算器能解决这一个问题, 但其中的碎片是可以计算的。 请在下面尝试一个, 或者输入您自己的 。
实践 (40)
先在纸上试一个。 将答案显示为检查; 校验的答案可以在求解器中打开, 每一步都可用 。
-
Define \(R\) as the region bounded above by the graph of \(f(x)=1\text{/}x\) and below by the \(x\text{-axis}\) over the interval \([1,3].\) Find the volume of the solid of revolution formed by revolving \(R\) around the \(y\text{-axis}.\)
发送答案
First we must graph the region \(R\) and the associated solid of revolution, as shown in the following figure.
Then the volume of the solid is given by
\[\begin{array}{ll}V & ={\int }_{a}^{b}(2\pi xf(x))dx \\ & ={\int }_{1}^{3}(2\pi x(\frac{1}{x}))dx \\ & ={\int }_{1}^{3}2\pi \ dx={2\pi x|}_{1}^{3}=4\pi \ {\text{units}}^{3}\text{.}\end{array}\] -
Define R as the region bounded above by the graph of \(f(x)={x}^{2}\) and below by the x-axis over the interval \([1,2].\) Find the volume of the solid of revolution formed by revolving \(R\) around the \(y\text{-axis}.\)
发送答案
\(\frac{15\pi }{2}\) units3
-
Define R as the region bounded above by the graph of \(f(x)=2x-{x}^{2}\) and below by the \(x\text{-axis}\) over the interval \([0,2].\) Find the volume of the solid of revolution formed by revolving \(R\) around the \(y\text{-axis}.\)
发送答案
First graph the region \(R\) and the associated solid of revolution, as shown in the following figure.
Then the volume of the solid is given by
\[\begin{array}{ll}V & ={\int }_{a}^{b}(2\pi xf(x))dx \\ & ={\int }_{0}^{2}(2\pi x(2x-{x}^{2}))dx=2\pi {\int }_{0}^{2}(2{x}^{2}-{x}^{3})dx \\ & ={2\pi [\frac{2{x}^{3}}{3}-\frac{{x}^{4}}{4}]\ |}_{0}^{2}=\frac{8\pi }{3}\ {\text{units}}^{3}\text{.}\end{array}\] -
Define \(R\) as the region bounded above by the graph of \(f(x)=3x-{x}^{2}\) and below by the \(x\text{-axis}\) over the interval \([0,2].\) Find the volume of the solid of revolution formed by revolving \(R\) around the \(y\text{-axis}.\)
发送答案
\(8\pi\) units3
-
Define \(Q\) as the region bounded on the right by the graph of \(g(y)=2\sqrt{y}\) and on the left by the \(y\text{-axis}\) for \(y\in [0,4].\) Find the volume of the solid of revolution formed by revolving \(Q\) around the x-axis.
发送答案
First, we need to graph the region \(Q\) and the associated solid of revolution, as shown in the following figure.
Label the shaded region \(Q.\) Then the volume of the solid is given by
\[\begin{array}{ll}V & ={\int }_{c}^{d}(2\pi yg(y))dy \\ & ={\int }_{0}^{4}(2\pi y(2\sqrt{y}))dy=4\pi {\int }_{0}^{4}{y}^{3\text{/}2}dy \\ & ={4\pi [\frac{2{y}^{5\text{/}2}}{5}]\ |}_{0}^{4}=\frac{256\pi }{5}\ {\text{units}}^{3}\text{.}\end{array}\] -
Define \(Q\) as the region bounded on the right by the graph of \(g(y)=3\text{/}y\) and on the left by the \(y\text{-axis}\) for \(y\in [1,3].\) Find the volume of the solid of revolution formed by revolving \(Q\) around the \(x\text{-axis}.\)
发送答案
\(12\pi\) units3
-
Define \(R\) as the region bounded above by the graph of \(f(x)=x\) and below by the \(x\text{-axis}\) over the interval \([1,2].\) Find the volume of the solid of revolution formed by revolving \(R\) around the line \(x=-1.\)
发送答案
First, graph the region \(R\) and the associated solid of revolution, as shown in the following figure.
Note that the radius of a shell is given by \(x+1.\) Then the volume of the solid is given by
\[\begin{array}{ll}V & ={\int }_{1}^{2}(2\pi (x+1)f(x))dx \\ & ={\int }_{1}^{2}(2\pi (x+1)x)dx=2\pi {\int }_{1}^{2}({x}^{2}+x)dx \\ & ={2\pi [\frac{{x}^{3}}{3}+\frac{{x}^{2}}{2}]\ |}_{1}^{2}=\frac{23\pi }{3}\ {\text{units}}^{3}\text{.}\end{array}\] -
Define \(R\) as the region bounded above by the graph of \(f(x)={x}^{2}\) and below by the \(x\text{-axis}\) over the interval \([0,1].\) Find the volume of the solid of revolution formed by revolving \(R\) around the line \(x=-2.\)
发送答案
\(\frac{11\pi }{6}\) units3
-
Define \(R\) as the region bounded above by the graph of the function \(f(x)=\sqrt{x}\) and below by the graph of the function \(g(x)=1\text{/}x\) over the interval \([1,4].\) Find the volume of the solid of revolution generated by revolving \(R\) around the \(y\text{-axis}.\)
发送答案
First, graph the region \(R\) and the associated solid of revolution, as shown in the following figure.
Note that the axis of revolution is the \(y\text{-axis},\) so the radius of a shell is given simply by \(x.\) We don’t need to make any adjustments to the x-term of our integrand. The height of a shell, though, is given by \(f(x)-g(x),\) so in this case we need to adjust the \(f(x)\) term of the integrand. Then the volume of the solid is given by
\[\begin{array}{ll}V & ={\int }_{1}^{4}(2\pi x(f(x)-g(x)))dx \\ & ={\int }_{1}^{4}(2\pi x(\sqrt{x}-\frac{1}{x}))dx=2\pi {\int }_{1}^{4}({x}^{3\text{/}2}-1)dx \\ & ={2\pi [\frac{2{x}^{5\text{/}2}}{5}-x]\ |}_{1}^{4}=\frac{94\pi }{5}\ {\text{units}}^{3}.\end{array}\] -
Define \(R\) as the region bounded above by the graph of \(f(x)=x\) and below by the graph of \(g(x)={x}^{2}\) over the interval \([0,1].\) Find the volume of the solid of revolution formed by revolving \(R\) around the \(y\text{-axis}.\)
发送答案
\(\frac{\pi }{6}\) units3
-
For each of the following problems, select the best method to find the volume of a solid of revolution generated by revolving the given region around the \(x\text{-axis},\) and set up the integral to find the volume (do not evaluate the integral).
- The region bounded by the graphs of \(y=x,\) \(y=2-x,\) and the \(x\text{-axis}.\)
- The region bounded by the graphs of \(y=4x-{x}^{2}\) and the \(x\text{-axis}.\)
发送答案
- First, sketch the region and the solid of revolution as shown.
Looking at the region, if we want to integrate with respect to \(x,\) we would have to break the integral into two pieces, because we have different functions bounding the region over \([0,1]\) and \([1,2].\) In this case, using the disk method, we would have
\[V={\int }_{0}^{1}(\pi {x}^{2})dx+{\int }_{1}^{2}(\pi {(2-x)}^{2})dx.\]
If we used the shell method instead, we would use functions of \(y\) to represent the curves, producing
\[\begin{array}{ll}V & ={\int }_{0}^{1}(2\pi y[(2-y)-y])dy \\ & ={\int }_{0}^{1}(2\pi y[2-2y])dy.\end{array}\]
Neither of these integrals is particularly onerous, but since the shell method requires only one integral, and the integrand requires less simplification, we should probably go with the shell method in this case. - First, sketch the region and the solid of revolution as shown.
Looking at the region, it would be problematic to define a horizontal rectangle; the region is bounded on the left and right by the same function. Therefore, we can dismiss the method of shells. The solid has no cavity in the middle, so we can use the method of disks. Then
\[V={\int }_{0}^{4}\pi {(4x-{x}^{2})}^{2}dx.\]
-
Select the best method to find the volume of a solid of revolution generated by revolving the given region around the \(x\text{-axis},\) and set up the integral to find the volume (do not evaluate the integral): the region bounded by the graphs of \(y=2-{x}^{2}\) and \(y={x}^{2}.\)
发送答案
Use the method of washers; \(V={\int }_{-1}^{1}\pi [{(2-{x}^{2})}^{2}-{({x}^{2})}^{2}]dx\)
-
[T] Bounded by the curves \(y=3x,x=0,\) and \(y=3\) rotated around the \(y\text{-axis}.\)
-
[T] Bounded by the curves \(y=3x,y=0,\ \text{and}\ x=3\) rotated around the \(y\text{-axis}.\)
发送答案
\(54\pi\) units3
-
[T] Bounded by the curves \(y=3x,y=0,\ \text{and}\ y=3\) rotated around the \(x\text{-axis}.\)
-
[T] Bounded by the curves \(y=3x,y=0,\ \text{and}\ x=3\) rotated around the \(x\text{-axis}.\)
发送答案
\(81\pi\) units3
-
[T] Bounded by the curves \(y=2{x}^{3},y=0,\ \text{and}\ x=2\) rotated around the \(y\text{-axis}.\)
-
[T] Bounded by the curves \(y=2{x}^{3},y=0,\ \text{and}\ x=2\) rotated around the \(x\text{-axis}.\)
发送答案
\(\frac{512\pi }{7}\) units3
-
\(y=1-{x}^{2},x=0,\ \text{and}\ x=1\)
-
\(y=5{x}^{3},x=0,\ \text{and}\ x=1\)
发送答案
\(2\pi\) units3
-
\(y=\frac{1}{x},x=1,\ \text{and}\ x=100\)
-
\(y=\sqrt{1-{x}^{2}},x=0,\ \text{and}\ x=1\)
发送答案
\(\frac{2\pi }{3}\) units3
-
\(y=\frac{1}{1+{x}^{2}},x=0,\ \text{and}\ x=3\)
-
\(y=\text{sin}{x}^{2},x=0,\ \text{and}\ x=\sqrt{\pi }\)
发送答案
\(2\pi\) units3
-
\(y=\frac{1}{\sqrt{1-{x}^{2}}},x=0,\ \text{and}\ x=\frac{1}{2}\)
-
\(y=\sqrt{x},x=0,\ \text{and}\ x=1\)
发送答案
\(\frac{4\pi }{5}\) units3
-
\(y={(1+{x}^{2})}^{3},x=0,\ \text{and}\ x=1\)
-
\(y=5{x}^{3}-2{x}^{4},x=0,\ \text{and}\ x=2\)
发送答案
\(\frac{64\pi }{3}\) units3
-
\(y=\sqrt{1-{x}^{2}},x=0,\ x=1\) and the x-axis
-
\(y={x}^{2},x=0,\ x=2\) and the x-axis
发送答案
\(\frac{32\pi }{5}\) units3
-
\(y=\frac{{x}^{3}}{2},\ x=0,\ x=2,\) and the x-axis
-
\(y=\frac{2}{{x}^{2}},\ x=1,\ x=2,\) and the x-axis
发送答案
\(\frac{7\pi }{6}\)
-
\(x=\frac{1}{1+{y}^{2}},y=4\)
-
\(x=\frac{1+{y}^{2}}{y},y=1,\ y=4,\) and the y-axis
发送答案
\(48\pi\)
-
\(x=\sqrt{4-{y}^{2}}\text{,}x=0\text{,}y=0\)
-
\(x={y}^{3}-2{y}^{2},\ x=0,\ x=9\)
发送答案
\(\frac{308\pi }{5}\)
-
\(x=\sqrt{y}+1,\ x=1,\ x=3,\) and the x-axis
-
\(x=\sqrt[3]{27y}\text{and}\ x=\frac{3y}{4}\)
发送答案
\(\frac{512\pi }{7}\)
-
\(y=3-x,y=0,x=0,\ \text{and}\ x=2\) rotated around the \(y\text{-axis}.\)
-
\(y={x}^{3},x=0,\ \text{and}\ y=8\) rotated around the \(y\text{-axis}.\)
发送答案
\(\frac{8\pi }{5}\)
此处所用符号
输入完整定义、图片和其中每个字母的含义的符号。
如何: Volumes of Revolution: Cylindrical Shells
- Calculate the volume of a solid of revolution by using the method of cylindrical shells.
- Compare the different methods for calculating a volume of revolution.
- The region bounded by the graphs of
- The region bounded by the graphs of
- First, sketch the region and the solid of revolution as shown.
- First, sketch the region and the solid of revolution as shown.
- The geometry of the functions and the difficulty of the integration are the main factors in deciding which integration method to use.
人们对人们提出的问题
What is a derivative in one sentence?
The slope of the graph at a point, the rate at which the output is changing there. Speed is the derivative of position.
What is an integral in one sentence?
The accumulated total of a rate: the area under the curve. Distance travelled is the integral of speed.
Why are derivatives and integrals opposites?
That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.
When do I use substitution and when integration by parts?
Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function, a polynomial times an exponential, log or trig function.
本页部分内容从 OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). 在这里进行浓缩和重新解释;错误属于我们。 由于其来源是非商业性用途许可证,这一课程是免费的,不出售任何物品,根据同一许可证分享。
更多 Calculus
LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests