Volumes

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 6.2 – Volumes
• Section 6.3 – Volumes by Cylindrical Shells

Expected Educational Results

• Objective 11–1: I can set up the integral to compute the volume obtained by rotating a region about an axis.
• Objective 11–2: I can evaluate the integral to compute the volume obtained by rotating a region about an axis using FTC-II.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Volumes

Complete Examples

Example 01: Washer Method – Horizontal Axis of Rotation

Question 01: Find the volume formed by rotating the region bounded by $f(x)=x$$f(x)=x$ and $f(x)=x^2$$f(x)=x^2$ about $x$$x$-axis.

Answer: Here is the sketch of the bounded region and showing

• one (1) typical rectangle in the region;
• the length for the radius function, $R(x)$$R(x)$;
• the length for the radius function, $r(x)$$r(x)$;
• the axis of rotation ($x$$x$-axis).

The formula to compute volume using Washer Method is:

${\displaystyle V=\pi {\int }_{x=a}^{x=b}[{(R(x))}^2-{(r(x))}^2]\text{d}x}$$\displaystyle V=\pi\int_{x=a}^{x=b}{\left[\left(R(x)\right)^2-\left(r(x)\right)^2\right]\,\text{d}x}$

where

• $R(x)$$R(x)$ is the outer radius of the washer (i.e., the vertical distance from the axis of rotation to the furthest side of the rectangle). To find the distance, subtract the smaller value from the larger value;
• $r(x)$$r(x)$ is the inner radius of the washer (i.e., the vertical distance from the axis of rotation to the closest side of the rectangle).

${\displaystyle V=\pi {\int }_{x=a}^{x=b}[{(R(x))}^2-{(r(x))}^2]\text{d}x=\pi {\int }_0^1[{\left(x\right)}^2-{\left(x^2\right)}^2]\text{d}x}$$\displaystyle V=\pi\int_{x=a}^{x=b}{\left[\left(R(x)\right)^2-\left(r(x)\right)^2\right]\,\text{d}x}=\pi\int_0^1{\left[\left(x\right)^2-\left(x^2\right)^2\right]\,\text{d}x}$

${\displaystyle \Rightarrow \pi {\int }_0^1[x^2-x^4]dx=\pi {(\frac{x^3}{3}-\frac{x^5}{5})}_0^1=\frac{2\pi }{15}}$$\displaystyle \Rightarrow\pi\int_0^1{\left[x^2-x^4\right]\,dx}=\pi\left(\frac{x^3}{3}-\frac{x^5}{5}\right)_0^1=\frac{2\pi}{15}$

Example 02: Washer Method – Vertical Axis of Rotation

Question 02: Find the volume formed by rotating the region bounded by $x=y^2-2y$$x=y^2-2y$ and $x=2-y$$x=2-y$ about $x=4$$x=4$.

Answer: Here is the sketch of the bounded region and showing

• one (1) typical rectangle in the region;
• the length for the radius function, $R(y)$$R(y)$;
• the length for the radius function, $r(y)$$r(y)$;
• the axis of rotation (dashed purple line).

The formula to compute volume using Washer Method is:

${\displaystyle V=\pi {\int }_{y=c}^{y=d}[{(R(y))}^2-{(r(y))}^2]\text{d}y}$$\displaystyle V=\pi\int_{y=c}^{y=d}{\left[\left(R(y)\right)^2-\left(r(y)\right)^2\right]\,\text{d}y}$

where

• $R(y)$$R(y)$ is the outer radius of the washer (i.e., the horizontal distance from the axis of rotation to the furthest side of the rectangle). To find the distance, subtract the smaller (left) value from the larger (right) value; ${\displaystyle R(y)=4-(y^2-2y)=4-y^2+2y}$$\displaystyle R(y)=4-(y^2-2y)=4-y^2+2y$.
• $r(y)$$r(y)$ is the inner radius of the washer (i.e., the horizontal distance from the axis of rotation to the closest side of the rectangle). To find the distance, subtract the smaller (left) value from the larger (right) value; ${\displaystyle r(y)=4-(2-y)=2+y}$$\displaystyle r(y)=4-(2-y)=2+y$

${\displaystyle V=\pi {\int }_{y=c}^{y=d}[{(R(y))}^2-{(r(y))}^2]\text{d}y=\pi {\int }_{-1}^2[{(4-y^2+2y)}^2-{(2+y)}^2]\text{d}y}$$\displaystyle V=\pi\int_{y=c}^{y=d}{\left[\left(R(y)\right)^2-\left(r(y)\right)^2\right]\,\text{d}y}=\pi\int_{-1}^2{\left[\left(4-y^2+2y\right)^2-\left(2+y\right)^2\right]\,\text{d}y}$

${\displaystyle \Rightarrow \pi {\int }_{-1}^2[y^4-4y^3-5y^2+12y+12]\text{d}y=\pi {(\frac{y^5}{5}-y^4-\frac{5y^3}{3}+6y^2+12y)}_{-1}^2=\frac{153\pi }{5}}$$\displaystyle \Rightarrow\pi\int_{-1}^2{\left[y^4-4 y^3-5 y^2+12 y+12\right]\,\text{d}y}=\pi\left(\frac{y^5}{5}-y^4-\frac{5y^3}{3}+6y^2+12y\right)_{-1}^{2}=\frac{153\pi}{5}$

Example 03: Shell Method – Vertical Axis of Rotation

Question 03: Find the volume formed by rotating the region bounded by $x=y^2-2y$$x=y^2-2y$ and $x=0$$x=0$ about $x$$x$-axis.

Answer: Here is the sketch of the bounded region and showing

• one (1) typical rectangle in the region;
• the length for the radius function, $p(y)$$p(y)$;
• the height of rectangle (i.e., length of long side) function, $h(y)$$h(y)$;
• the axis of rotation ($y$$y$-axis).

The formula to compute volume using Shell Method is:

${\displaystyle V=2\pi {\int }_{y=c}^{y=d}h(y)p(y)\text{d}y}$$\displaystyle V=2\pi\int_{y=c}^{y=d}{h(y)p(y)\,\text{d}y}$

where

• $h(y)$$h(y)$ is the height of the shell (i.e., the horizontal length of the long side of the rectangle) found by subtracting the lower (left) function from the upper (right) function; ${\displaystyle h(y)=0-(y^2-2y)=2y-y^2}$$\displaystyle h(y)=0-(y^2-2y)=2y-y^2$.
• $p(y)$$p(y)$ is the radius of the shell (i.e., the vertical distance the rectangle is from the axis of rotation); since the rectangle is located $y$$y$-units vertically from the $x$$x$-axis, $p(y)=y$$p(y)=y$.

${\displaystyle V=2\pi {\int }_0^{2}h(y)p(y)\text{d}y=2\pi {\int }_0^2(2y-y^2)y\text{d}y=2\pi {\int }_0^2(2y^2-y^3)\text{d}y}$$\displaystyle V=2\pi\int_0^2{h(y)p(y)\,\text{d}y}=2\pi\int_0^2{\left(2y-y^2\right)y\,\text{d}y}=2\pi\int_0^2{\left(2y^2-y^3\right)\,\text{d}y}$

${\displaystyle \Rightarrow 2\pi {(\frac{2y^3}{3}-\frac{y^4}{4})}_0^2=\frac{8\pi }{3}}$$\displaystyle \Rightarrow 2\pi\left(\frac{2y^3}{3}-\frac{y^4}{4}\right)_0^2=\frac{8\pi}{3}$

Example 04: Shell Method – Horizontal Axis of Rotation

Question 04: Find the volume formed by rotating the region bounded by $y=x^3+x+1$$y=x^3+x+1$, $x=1$$x=1$ and $y=1$$y=1$ about $x=2$$x=2$.

Answer: Here is the sketch of the bounded region and showing

• one (1) typical rectangle in the region;
• the length for the radius function, $p(x)$$p(x)$;
• the height of rectangle (i.e., length of long side) function, $h(x)$$h(x)$;
• the axis of rotation (purple dashed line).

The formula to compute volume using Shell Method is:

${\displaystyle V=2\pi {\int }_{x=a}^{x=b}h(x)p(x)\text{d}x}$$\displaystyle V=2\pi\int_{x=a}^{x=b}{h(x)p(x)\,\text{d}x}$

where

• $h(x)$$h(x)$ is the height of the shell (i.e., the vertical length of the long side of the rectangle) found by subtracting the lower function from the upper function; ${\displaystyle h(x)=(x^3+x+1)-1=x^3+x}$$\displaystyle h(x)=(x^3+x+1)-1=x^3+x$.
• $p(x)$$p(x)$ is the radius of the shell (i.e., the horizontal distance the rectangle is from the axis of rotation); since the rectangle is located $x$$x$-units vertically from the $y$$y$-axis, $p(x)=2-x$$p(x)=2-x$.

${\displaystyle V=2\pi {\int }_0^{1}h(x)p(x)\text{d}x=2\pi {\int }_0^1(x^3+x)(2-x)\text{d}x=2\pi {\int }_0^1(-x^4+2x^3-x^2+2x)\text{d}x}$$\displaystyle V=2\pi\int_0^1{h(x)p(x)\,\text{d}x}=2\pi\int_0^1{\left(x^3+x\right)\left(2-x\right)\,\text{d}x}=2\pi\int_0^1{\left(-x^4+2x^3-x^2+2x\right)\,\text{d}x}$

${\displaystyle \Rightarrow {(-\frac{x^5}{5}+\frac{x^4}{2}-\frac{x^3}{3}+x^2)}_0^1=\frac{29\pi }{15}}$$\displaystyle \Rightarrow \left(-\frac{x^5}{5}+\frac{x^4}{2}-\frac{x^3}{3}+x^2\right)_0^1=\frac{29\pi}{15}$

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Created: Wednesday, 02 March 2022 - 02:17 (EST)