Volume Exercises

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Here we’ll practice finding volume using integration.

Consider the region bounded by $y = \frac {5x}{2}-x^2$ and $y=\frac {x}{2}$ . What is the volume of the solid obtained by revolving this region about the $x$ -axis?

$\textrm {Volume} = \answer [given]{\frac {12 \pi }{5}}$

Consider the region bounded by $y =\sqrt {x-1}$ , the $x$ -axis, and the vertical line $x=2$ . What is the volume of the solid obtained by revolving this region about the $y$ -axis?

Draw a picture!

Solving for $x$ , we have $x = y^2+1$ . Note that $y$ ranges from $0$ to $1$ as $x$ goes from $1$ to $2$ .

We can decompose the solid into infinitesmal washers with width $\d y$ , inner radius $y^2+1$ and outer radius $2$ . The volume of each washer is $\pi ((2)^2 - (y^2+1)^2)\d y$ . Summing these volumes from $y=0$ to $y=1$ , we obtain $\textrm {Volume} = \int _0^1 \pi ((2)^2 - (y^2+1)^2)\d y$

$\begin{align*} \int_0^1\pi((2)^2 - (y^2+1)^2)\d y &= \int_0^1 \pi(3-2y^2-y^4)\d y\\ &= \pi \eval{\frac{-y^5}{5}+\frac{-2y^3}{3}+3y}_0^1\\ &=\pi(\frac{-1}{5}+\frac{-2}{3}+3)\\ &=\frac{32\pi}{15} \end{align*}$

$\textrm {Volume} = \answer {\frac {32\pi }{15}}$

Consider the region bounded by $y =x^2$ and $y=4$ . A solid has this region as its base, and the cross sections of the solid when cut with planes parallel to the $(y,z)$ -plane are all squares. What is the area of the solid?

Draw a picture! Note that the intersections occur at $x= \pm 2$

We can decompose the solid into square slabs with width $\d x$ and side lengths $4-x^2$ . The volume of each slab is $(4-x^2)\d x$ . Summing these volumes from $x=-2$ to $x=2$ , we obtain $\textrm {Volume} = \int _{-2}^{2} (4-x^2)^2\d x$

First note that this function is even, so we may use symmetry to rewrite the integral as $2\int _{0}^{2} (4-x^2)^2\d x$ $\begin{align*} 2\int_{0}^{2} (4-x^2)^2\d x &=2 \int_{0}^{2} 16-8x^2+x^4 \d x \\ &=2 \eval{16x-\frac{8x^3}{3}+\frac{x^5}{5}}_0^2 \\ &=2(32-\frac{64}{3}+\frac{32}{5})\\ &=\frac{512}{15} \end{align*}$

$\textrm {Volume} = \answer {\frac {512}{15}}$

Consider the region bounded by $y = x$ and $y=\frac {x^3}{4}$ in the first quadrant. What is the volume of the solid obtained by revolving this region about the line $y=-1$ ?

Draw a picture!

First we find the points of intersectios. $\begin{align*} x&= \frac{x^3}{4}\\ 4x &= x^3\\ x^3-4x &=0\\ x(x-2)(x+2) &=0 \end{align*}$

So the points of intersection are $x=-2$ , $x=0$ , $x=2$ . Since we only care about the first quadrant, our bounds are from $x=0$ to $x=2$ .

By graphing the two functions, we can see that $y=x$ is always greater than $y=\frac {x^3}{4}$ on the interval $[0,2]$ .

We can decompose the solid into infinitesmal washers with width $dx$ , inner radius $\frac {x^3}{4}+1$ and outer radius $x+1$ . The volume of each washer is $\pi ((x+1)^2-(\frac {x^3}{4}+1)^2)\d x$ . Summing these volumes from $x=0$ to $x=2$ , we obtain

$\textrm {Volume} = \int _0^2 \pi (x+1)^2-(\frac {x^3}{4}+1)^2) \d x$

By expanding this polynomial, we find that this evaluates to $\frac {74\pi }{21}$

$\textrm {Volume} = \answer {\frac {74\pi }{21}}$

Consider the region bounded by the lines $y = x$ , $y=\frac {x}{2}$ , $y=1$ , and $y = 4$ . What is the volume of the solid obtained by revolving this region about the line $x=-2$ ?

Draw a picture!

Solving for $x$ , we have $x=y$ and $x = 2y$ .

We can decompose the solid into infinitesmal washers with width $\d y$ , inner radius $y+2$ and outer radius $2y+2$ . The volume of each washer is $\pi ((2y+2)^2-(y+2)^2)\d y$ . Summing these volumes from $y=1$ to $y=4$ , we obtain $\textrm {Volume} = \int _1^4 \pi ((2y+2)^2-(y+2)^2)\d y$

By expanding this polynomial, we find that this evaluates to $93\pi$ .

$\textrm {Volume} = \answer {93\pi }$

Consider the region bounded by the lines $y = \sin (x)$ , $x=0$ , $x=\pi$ , and the $x$ -axis. What is the volume of the solid obtained by revolving this region about $x$ -axis?

It will be useful to recall that $\sin ^2(x) = \frac {1-\cos (2x)}{2}$ .

Draw a picture!

We can decompose the solid into infinitesmal disks with width $\d x$ and radius $\sin (x)$ . The volume of each washer is $\pi \sin ^2(x)\d x$ . Summing these volumes from $x=0$ to $x=\pi$ , we obtain $\textrm {Volume} = \int _0^\pi \pi \sin ^2(x)\d x$

Using the half angle reduction formula, and a substitution, we obtain that this evaluates to $\frac {\pi ^2}{2}$

$\textrm {Volume} = \answer {\frac {\pi ^2}{2}}$

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