# Determining the Volume of a Solid of Revolution

Volumes of Revolution - Disk/Washers Example 1
4:35 — by patrickJMT

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• The volume of a solid of revolution for a curve $y = f \left(x\right)$ rotated around the $x$-axis can be found through: $V = \pi {\int}_{a}^{b} {y}^{2} \mathrm{dx}$
To solve this integral you must find the bounds $a$ and $b$.

For example, the curve $y = {x}^{2} + 2$ is rotated between the bounds $x = 1$ and $x = 2$ is rotated about the $x$-axis. Find the volume of the solid formed.
Since $y = {x}^{2} + 2$
${y}^{2} = {x}^{4} + 4 {x}^{2} + 4$
$\therefore V = \pi {\int}_{1}^{2} {x}^{4} + 4 {x}^{2} + 4 \mathrm{dx}$
$= \pi {\left[\frac{{x}^{5}}{5} + \frac{4 {x}^{3}}{3} + 4 x\right]}_{1}^{2}$
$= \frac{431 \pi}{5}$ or $270.81$(2dp) units cubed

When the curve is rotated around the $y$-axis the volume of the solid is given by: $V = \pi {\int}_{e}^{d} {x}^{2} \mathrm{dx}$
The function must be in the form where $x$ is a function of $y$ and the bounds $d$ and $e$ must be read off the $y$-axis.

• The short and sweet version of the answer is:
$\left(\pi\right) \cdot {\int}_{0}^{1} \left({\left(4 - {x}^{2}\right)}^{2} - {\left(4 - x\right)}^{2}\right) \mathrm{dx}$=

$1.2 \pi$ or approximately $3.769911 . . .$

This is the region bounded by the points $\left(0 , 0\right)$ and $\left(1 , 1\right)$ since those points lie on both the graph of the line ($g \left(x\right) = x$) and the parabola ((f(x)=x^2). We will take this region and subdivide it into vertical regions that are one can picture as tiny rectangle. When the rectangles are spun about the line $y = 4$, they will form tiny washers. If there are $n$ subintervals, then the width of each subinterval is $\Delta x = \frac{1}{n}$. The outer radius (distance from the function farther away from the line of rotation to the line of rotation) is $\left(4 - f \left(x\right)\right)$. The inner radius (distance from the function nearer to the line of rotation to the line of rotation) is (4-g(x))

The usual setup for washers is:

$\cdot {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \left({\left(\pi R \left(x\right)\right)}^{2} - {\left(\pi r \left(x\right)\right)}^{2}\right) \cdot \left(\Delta x\right)$ which becomes

(pi)*int_a^b((piR(x)^2-(pir(x))^2)*(dx)# which for our functions equals:

${\int}_{0}^{1} \left(\pi {\left(4 - f \left(x\right)\right)}^{2} - \pi {\left(4 - g \left(x\right)\right)}^{2}\right) \mathrm{dx}$=$\left(\pi\right) \cdot {\int}_{0}^{1} \left({\left(4 - {x}^{2}\right)}^{2} - {\left(4 - x\right)}^{2}\right) \mathrm{dx}$=

$1.2 \pi$ or approximately $3.769911 . . .$

• A cone with base radius $r$ and height $h$ can be obtained by rotating the region under the line $y = \frac{r}{h} x$ about the x-axis from $x = 0$ to $x = h$.
By Disk Method,
$V = \pi {\int}_{0}^{h} {\left(\frac{r}{h} x\right)}^{2} \mathrm{dx} = \frac{\pi {r}^{2}}{{h}^{2}} {\int}_{0}^{h} {x}^{2} \mathrm{dx}$
by Power Rule,
$= \frac{\pi {r}^{2}}{h} ^ 2 {\left[{x}^{3} / 3\right]}_{0}^{h} = \frac{\pi {r}^{2}}{{h}^{2}} \cdot {h}^{3} / 3 = \frac{1}{3} \pi {r}^{2} h$

• If the radius of its circular cross section is $r$, and the radius of the circle traced by the center of the cross sections is $R$, then the volume of the torus is $V = 2 {\pi}^{2} {r}^{2} R$.

Let's say the torus is obtained by rotating the circular region ${x}^{2} + {\left(y - R\right)}^{2} = {r}^{2}$ about the $x$-axis. Notice that this circular region is the region between the curves: $y = \sqrt{{r}^{2} - {x}^{2}} + R$ and $y = - \sqrt{{r}^{2} - {x}^{2}} + R$.

By Washer Method, the volume of the solid of revolution can be expressed as:
$V = \pi {\int}_{- r}^{r} \left[{\left(\sqrt{{r}^{2} - {x}^{2}} + R\right)}^{2} - {\left(- \sqrt{{r}^{2} - {x}^{2}} + R\right)}^{2}\right] \mathrm{dx}$,
which simplifies to:
$V = 4 \pi R \setminus {\int}_{- r}^{r} \sqrt{{r}^{2} - {x}^{2}} \mathrm{dx}$
Since the integral above is equivalent to the area of a semicircle with radius r, we have
$V = 4 \pi R \cdot \frac{1}{2} \pi {r}^{2} = 2 {\pi}^{2} {r}^{2} R$

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