We will be focused on the volumes of solids obtained by revolving a region around an axis. Let’s use the running example of the region bounded by the curves \(x=0,y=4,y=x^2\text{.}\)
(a)
Consider the below illustrated revolution of this region, and the cross-section drawn from a horizontal line segment. Choose the most appropriate description of this illustration.
Region is rotated around the \(x\)-axis; the cross-sectional area is determined by the line segment’s \(x\)-value.
Region is rotated around the \(x\)-axis; the cross-sectional area is determined by the line segment’s \(y\)-value.
Region is rotated around the \(y\)-axis; the cross-sectional area is determined by the line segment’s \(x\)-value.
Region is rotated around the \(y\)-axis; the cross-sectional area is determined by the line segment’s \(y\)-value.
(b)
Which of these formulas is most appropriate to find this illustration’s cross-sectional area?
\(\displaystyle \pi r^2\)
\(\displaystyle 2\pi rh\)
\(\displaystyle \pi R^2-\pi r^2\)
\(\displaystyle \frac{1}{2}bh\)
(c)
Consider the below illustrated revolution of this region, and the cross-section drawn from a vertical line segment. Choose the most appropriate description of this illustration.
Region is rotated around the \(x\)-axis; the cross-sectional area is determined by the line segment’s \(x\)-value.
Region is rotated around the \(x\)-axis; the cross-sectional area is determined by the line segment’s \(y\)-value.
Region is rotated around the \(y\)-axis; the cross-sectional area is determined by the line segment’s \(x\)-value.
Region is rotated around the \(y\)-axis; the cross-sectional area is determined by the line segment’s \(y\)-value.
(d)
Which of these formulas is most appropriate to find this illustration’s cross-sectional area?
\(\displaystyle \pi r^2\)
\(\displaystyle 2\pi rh\)
\(\displaystyle \pi R^2-\pi r^2\)
\(\displaystyle \frac{1}{2}bh\)
(e)
Consider the below illustrated revolution of this region, and the cross-section drawn from a horizontal line segment. Choose the most appropriate description of this illustration.
Region is rotated around the \(x\)-axis; the cross-sectional area is determined by the line segment’s \(x\)-value.
Region is rotated around the \(x\)-axis; the cross-sectional area is determined by the line segment’s \(y\)-value.
Region is rotated around the \(y\)-axis; the cross-sectional area is determined by the line segment’s \(x\)-value.
Region is rotated around the \(y\)-axis; the cross-sectional area is determined by the line segment’s \(y\)-value.
(f)
Which of these formulas is most appropriate to find this illustration’s cross-sectional area?
\(\displaystyle \pi r^2\)
\(\displaystyle 2\pi rh\)
\(\displaystyle \pi R^2-\pi r^2\)
\(\displaystyle \frac{1}{2}bh\)
(g)
Consider the below illustrated revolution of this region, and the cross-section drawn from a vertical line segment. Choose the most appropriate description of this illustration.
Region is rotated around the \(x\)-axis; the cross-sectional area is determined by the line segment’s \(x\)-value.
Region is rotated around the \(x\)-axis; the cross-sectional area is determined by the line segment’s \(y\)-value.
Region is rotated around the \(y\)-axis; the cross-sectional area is determined by the line segment’s \(x\)-value.
Region is rotated around the \(y\)-axis; the cross-sectional area is determined by the line segment’s \(y\)-value.
(h)
Which of these formulas is most appropriate to find this illustration’s cross-sectional area?
\(\displaystyle \pi r^2\)
\(\displaystyle 2\pi rh\)
\(\displaystyle \pi R^2-\pi r^2\)
\(\displaystyle \frac{1}{2}bh\)
Remark6.3.4.
Generally when solving problems without the aid of technology, it’s useful to draw your region in two dimensions, choose whether to use a horizontal or vertical line segment, and draw its rotation to determine the cross-sectional shape.
When the shape is a disk, this is called the disk method and we use one of these formulas depending on whether the cross-sectional area depends on \(x\) or \(y\text{.}\)
When the shape is a washer, this is called the washer method and we use one of these formulas depending on whether the cross-sectional area depends on \(x\) or \(y\text{.}\)
When the shape is a cylindrical shell, this is called the shell method and we use one of these formulas depending on whether the cross-sectional area depends on \(x\) or \(y\text{.}\)
Let’s now consider the region bounded by the curves \(x=0,x=1,y=0,y=5e^x\text{,}\) rotated about the \(x\)-axis.
(a)
Sketch two copies of this region in the \(xy\) plane.
(b)
Draw a vertical line segment in one region and its rotation around the \(x\)-axis. Draw a horizontal line segment in the other region and its rotation around the \(x\)-axis.
(c)
Consider the method required for each cross-section drawn. Which would be the easiest strategy to proceed with?
The horizontal line segment, using the disk/washer method.
The horizontal line segment, using the shell method.
The vertical line segment, using the disk/washer method.
The vertical line segment, using the shell method.
(d)
Let’s proceed with the vertical segment. Which formula is most appropriate for the radius?
\(\displaystyle r(x)=x\)
\(\displaystyle r(x)=5e^x\)
\(\displaystyle r(x)=5\ln(x)\)
\(\displaystyle r(x)=\frac{1}{5}\ln(x)\)
(e)
Which of these integrals is equal to the volume of the solid of revolution?
\(\displaystyle \int_0^1 25\pi e^{2x}\,dx\)
\(\displaystyle \int_0^1 5\pi^2 e^{x}\,dx\)
\(\displaystyle \int_0^2 25\pi e^{x}\,dx\)
\(\displaystyle \int_0^2 5\pi^2 e^{2x}\,dx\)
Activity6.3.6.
Let’s now consider the same region, bounded by the curves \(x=0,x=1,y=0,y=5e^x\text{,}\) but this time rotated about the \(y\)-axis.
(a)
Sketch two copies of this region in the \(xy\) plane.
(b)
Draw a vertical line segment in one region and its rotation around the \(y\)-axis. Draw a horizontal line segment in the other region and its rotation around the \(y\)-axis.
(c)
Consider the method required for each cross-section drawn. Which would be the easiest strategy to proceed with?
The horizontal line segment, using the disk/washer method.
The horizontal line segment, using the shell method.
The vertical line segment, using the disk/washer method.
The vertical line segment, using the shell method.
(d)
Let’s proceed with the vertical segment. Which formula is most appropriate for the radius?
\(\displaystyle r(x)=x\)
\(\displaystyle r(x)=5e^x\)
\(\displaystyle r(x)=5\ln(x)\)
\(\displaystyle r(x)=\frac{1}{5}\ln(x)\)
(e)
Which formula is most appropriate for the height?
\(\displaystyle h(x)=x\)
\(\displaystyle h(x)=5e^x\)
\(\displaystyle h(x)=5\ln(x)\)
\(\displaystyle h(x)=\frac{1}{5}\ln(x)\)
(f)
Which of these integrals is equal to the volume of the solid of revolution?
\(\displaystyle \int_0^1 5\pi^2 xe^{x}\,dx\)
\(\displaystyle \int_0^1 10\pi xe^{x}\,dx\)
\(\displaystyle \int_0^2 5\pi xe^{x}\,dx\)
\(\displaystyle \int_0^2 10\pi x^2e^{x}\,dx\)
Activity6.3.7.
Consider the region bounded by \(y=2 \, x + 3, y=0, x=4, x=7\text{.}\)
(a)
Find an integral which computes the volume of the solid formed by rotating this region about the \(x\)-axis.
(b)
Find an integral which computes the volume of the solid formed by rotating this region about the \(y\)-axis.