# Thread: Volume and Surface Area of a Rotated Solid and Arclength

1. ## Volume and Surface Area of a Rotated Solid and Arclength

Hello, I'm back again. I have three problems (two with two parts) that, if you guys wouldn't mind, I would like to have checked. If anything is wrong, I would highly appreciate it if you showed me where I went wrong and how to do it correctly. Thanks infinitely in advance.

Problem 1.
Let R be the region bounded by the semicircle y = $\sqrt{r^2 - x^2}$ and the x-axis.

a. Derive the formula for the volume of a sphere of radius r by revolving R about the x-axis using the disk method or the shell method.

My Solution (Using Disk Method)

V = $\pi\int^r_0$ $r^2$ dx = $\pi[xr^2]^r_0$ = $\pi$ $r^3$

b. Derive the formula for the surface area of a sphere of radius r by revolving R about the x-axis.

A = $2\pi\int^r_0[\sqrt{r^2 - x^2}$ $\sqrt{1 - (\frac{x}{\sqrt{r^2 - x^2}})^2}]$ dx = ???

Problem 2. Let R be the region bounded by the lines x = 0, x = h, y = 0, and y = $\frac{r}{h}$x, where r and h are positive constants.

a. Derive the formula for the volume of a right circular cone of radius r and altitude h by revolving R about the x-axis using the disk method or the shell method.

My Solution (Using Disk Method)

$V = \pi\int^h_0(\frac{r}{h}x)^2$ dx = $\pi\int^h_0\frac{r^2}{h^2}x^2$ dx = $\pi[\frac{r^2}{3h^2}x^3]^h_0$ = $\frac{\pi(r^2)h}{3}$

b. Now derive the formula for the surface area of such a cone, again, by revolving R about the x-axis.

My Solution

A = $2\pi\int^h_0(\frac{r}{h}x)\sqrt{1 + x^2}$ dx = $2\pi\int^h_0(\frac{r}{h}x)(1 + x^2)^\frac{1}{2}$ dx = $2\pi[\frac{r}{3h}(1 + x^2)^\frac{3}{2}]^h_0$ = $2\pi[\frac{r}{3h}(1 + h^2)^\frac{3}{2}] - \frac{2\pi(r)}{3h}$ = $\frac{2\pi(r)}{3h}(1 + h^2)^\frac{3}{2} - \frac{2\pi(r)}{3h}$

Problem 3. Calculate the length along the curve y = 1 + $x^\frac{3}{2}$ from (0,1) to (4,9).

My Solution

L = $\int^4_0$ $\sqrt{1+(\frac{3}{2}\sqrt{x})^2}$ dx = $\int^4_0\sqrt{1 + \frac{9}{4}x}$ dx = $\int^4_0(1 + \frac{9}{4}x)^\frac{1}{2}$ dx = $[\frac{2}{3}(1 + \frac{9}{4}x)^\frac{3}{2}(\frac{4}{9})]^4_0$ = $[\frac{8}{27}(1 + \frac{9}{4}x)^\frac{3}{2}]^4_0$ = 8 - $\frac{8}{27}$ = $\frac{208}{27}$

Once again, thank you.

2. Hello mturner07
Originally Posted by mturner07
Hello, I'm back again. I have three problems (two with two parts) that, if you guys wouldn't mind, I would like to have checked. If anything is wrong, I would highly appreciate it if you showed me where I went wrong and how to do it correctly. Thanks infinitely in advance.

Problem 1.
Let R be the region bounded by the semicircle y = $\sqrt{r^2 - x^2}$ and the x-axis.

a. Derive the formula for the volume of a sphere of radius r by revolving R about the x-axis using the disk method or the shell method.

My Solution (Using Disk Method)

V = $\pi\int^r_0$ $r^2$ dx = $\pi[xr^2]^r_0$ = $\pi$ $r^3$
Thanks for showing us your working, but there are a few problems. Here's how you should start the first one.

The semicircle lies between $x = -r$ and $x = r$. When we divide the sphere into vertical discs, each disc has a radius $y$ and thickness $\delta x$. So the total volume is:
$\pi\int_{-r}^ry^2\;dx =\pi\int_{-r}^r(r^2-x^2)\;dx$
Do you want to have another attempt from here?

3. Hello mturner07

Here are some more suggestions.
Originally Posted by mturner07
...Problem 1. Let R be the region bounded by the semicircle y = $\sqrt{r^2 - x^2}$ and the x-axis.
...
b. Derive the formula for the surface area of a sphere of radius r by revolving R about the x-axis.

A = $2\pi\int^r_0[\sqrt{r^2 - x^2}$ $\sqrt{1 - (\frac{x}{\sqrt{r^2 - x^2}})^2}]$ dx = ???
Your integral is mostly correct: but it should be a $+$ sign, and you have the wrong limits again. Like this:
$A = 2\pi\int_{-r}^ry\sqrt{1+\left(\frac{dy}{dx}\right)^2}\;dx$
Now we have $y = \sqrt{r^2-x^2}$ and $\frac{dy}{dx}=-\frac{x}{y}$, and you'll find it's easier to write the integral like this before making the substitution:
$\Rightarrow A = 2\pi\int_{-r}^r\sqrt{y^2+y^2\left(\frac{dy}{dx}\right)^2}\;dx$
$=2\pi\int_{-r}^r\sqrt{y^2+y^2\left(-\frac{x}{y}\right)^2}\;dx$

$=2\pi\int_{-r}^r\left(\sqrt{y^2+x^2}\right)^2\;dx$

$=2\pi\int_{-r}^r\left(\sqrt{r^2}\right)^2\;dx$
I'm sure you can finish off from here.
Problem 2. Let R be the region bounded by the lines x = 0, x = h, y = 0, and y = $\frac{r}{h}$x, where r and h are positive constants.

a. Derive the formula for the volume of a right circular cone of radius r and altitude h by revolving R about the x-axis using the disk method or the shell method.

My Solution (Using Disk Method)

$V = \pi\int^h_0(\frac{r}{h}x)^2$ dx = $\pi\int^h_0\frac{r^2}{h^2}x^2$ dx = $\pi[\frac{r^2}{3h^2}x^3]^h_0$ = $\frac{\pi(r^2)h}{3}$
Correct! Well done!

b. Now derive the formula for the surface area of such a cone, again, by revolving R about the x-axis.

My Solution

A = $2\pi\int^h_0(\frac{r}{h}x)\sqrt{1 + x^2}$ dx = $2\pi\int^h_0(\frac{r}{h}x)(1 + x^2)^\frac{1}{2}$ dx = $2\pi[\frac{r}{3h}(1 + x^2)^\frac{3}{2}]^h_0$ = $2\pi[\frac{r}{3h}(1 + h^2)^\frac{3}{2}] - \frac{2\pi(r)}{3h}$ = $\frac{2\pi(r)}{3h}(1 + h^2)^\frac{3}{2} - \frac{2\pi(r)}{3h}$
Your initial integral is wrong. $\frac{dy}{dx} = \frac{r}{h}$, so it should be:
$A = 2\pi\int_0^h\frac{rx}{h}\sqrt{1+\frac{r^2}{h^2}}\; dx$
Most of the stuff inside the integral is constant, so it just ends up as:
$A = \frac{2\pi r}{h}\sqrt{1+\frac{r^2}{h^2}}\int_0^hx\;dx$
You can finish from here.
Problem 3. Calculate the length along the curve y = 1 + $x^\frac{3}{2}$ from (0,1) to (4,9).

My Solution

L = $\int^4_0$ $\sqrt{1+(\frac{3}{2}\sqrt{x})^2}$ dx = $\int^4_0\sqrt{1 + \frac{9}{4}x}$ dx = $\int^4_0(1 + \frac{9}{4}x)^\frac{1}{2}$ dx = $[\frac{2}{3}(1 + \frac{9}{4}x)^\frac{3}{2}(\frac{4}{9})]^4_0$ = $[\frac{8}{27}(1 + \frac{9}{4}x)^\frac{3}{2}]^4_0$ = 8 - $\frac{8}{27}$ = $\frac{208}{27}$
I think this is nearly all correct, but you've got the arithmetic wrong at the end.
$\left[\frac{8}{27}\left(1 + \frac{9}{4}x\right)^\frac{3}{2}\right]^4_0=\frac{8}{27}\left(10^{\frac32}-1\right)$

4. Thanks for the help!

With your help I get $\frac{4}{3}\pi(r)^3$ for problem 1a., $\pi(r)h\sqrt{1 + \frac{r^2}{h^2}}$ for problem 2b., and $\frac{80\sqrt{10} - 8}{27}$ for problem 3.

However, I have a question about your solution to problem 1a. Where does the square that squares $\sqrt{y^2 + x^2}$ come from in the step that has $2\pi\int^r_{-r}(\sqrt{y^2 + x^2})^2$ dx?

Once again, thank you!

5. Hello mturner07
Originally Posted by mturner07
Thanks for the help!

With your help I get $\frac{4}{3}\pi(r)^3$ for problem 1a., $\pi(r)h\sqrt{1 + \frac{r^2}{h^2}}$ for problem 2b., and $\frac{80\sqrt{10} - 8}{27}$ for problem 3.

However, I have a question about your solution to problem 1a. Where does the square that squares $\sqrt{y^2 + x^2}$ come from in the step that has $2\pi\int^r_{-r}(\sqrt{y^2 + x^2})^2$ dx?

Once again, thank you!
Sorry, that's a typo - it shouldn't be there! Should read
$2\pi\int_{-r}^r\left(\sqrt{y^2+x^2}\right)\;dx$

6. Alright. For problem 1a I get $\frac{4}{3}\pi(r)^3$, for problem 1b I get $4\pi(r)^2$, for problem 2b I get $\pi(r)h\sqrt{1 + \frac{r^2}{h^2}}$, and for problem 3 I get $\frac{80\sqrt{10} - 8}{27}$. Are these solutions consistent with yours? Thanks in advance.

7. Hello mturner07
Originally Posted by mturner07
Alright. For problem 1a I get $\frac{4}{3}\pi(r)^3$, for problem 1b I get $4\pi(r)^2$, for problem 2b I get $\pi(r)h\sqrt{1 + \frac{r^2}{h^2}}$, and for problem 3 I get $\frac{80\sqrt{10} - 8}{27}$. Are these solutions consistent with yours? Thanks in advance.
Yes, I agree with all these answers. The volume and surface area of a sphere are $\tfrac43\pi r^3$ and $4\pi r^2$ respectively.

You also have a correct expression for the curved surface area of a cone, which can then be simplified as follows:
$\pi rh\sqrt{1 + \frac{r^2}{h^2}}=\pi r\sqrt{h^2+r^2}$
$=\pi rl$
where $l$ is the 'slant-height' of the cone; i.e. the distance from a point on the circumference of the base to the vertex.

$\frac{8}{27}\left(10^{\frac32}-1\right) = \frac{8(10\sqrt{10}-1)}{27}$
$=\frac{80\sqrt{10}-8}{27}$