# Thread: Volume of a solid

1. ## Volume of a solid

I have this problem that wants me to find the volume of a solid. The problem is as follows:

Find the volume of the solid obtained by rotating the region bounded by $y = 0$, $y = (tan(x))^3sec(x)$, and $x = 1/3*pi$, about the x-axis.

I'm having trouble getting started with this problem. I've never done a problem like this and I can't find anything helpful in my textbook (I'm probably not looking in the right direction). I was wondering if anyone could give me some tips on how to get started or how to approach the problem.

Thanks for you consideration,

Austin Martin

2. Originally Posted by auslmar
I have this problem that wants me to find the volume of a solid. The problem is as follows:

Find the volume of the solid obtained by rotating the region bounded by $y = 0$, $y = (tan(x))^3sec(x)$, and $x = 1/3*pi$, about the x-axis.

I'm having trouble getting started with this problem. I've never done a problem like this and I can't find anything helpful in my textbook (I'm probably not looking in the right direction). I was wondering if anyone could give me some tips on how to get started or how to approach the problem.

Thanks for you consideration,

Austin Martin
Use the formula for volume about the x-axis

$V_{\text{revolution}}=\int_a^{b}f(x)^2dx$

So we would have assuming the left bound is the x-axis

$\int_0^{\frac{\pi}{3}}\tan^6(x)\sec^2(x)dx$

Then let $\varphi=\tan(x)$

3. Originally Posted by Mathstud28
Use the formula for volume about the x-axis

$V_{\text{revolution}}=\int_a^{b}f(x)^2dx$

So we would have assuming the left bound is the x-axis

$\int_0^{\frac{\pi}{3}}\tan^6(x)\sec^2(x)dx$

Then let $\varphi=\tan(x)$
Thanks for the help.

I do have a follow-up question, though.

I applied the formula you suggested and came up with

$\int_0^\frac{\pi}{3}tan^6(x)sec^2(x) = \frac{tan^7(\frac{\pi}{3})}{7}$ which is not the correct answer. Either I messed up integrating or the previously suggest approach is incorrect. Thanks in advance.

4. Hello,

Originally Posted by auslmar
Thanks for the help.

I do have a follow-up question, though.

I applied the formula you suggested and came up with

$\int_0^\frac{\pi}{3}tan^6(x)sec^2(x) = \frac{tan^7(\frac{\pi}{3})}{7}$ which is not the correct answer. Either I messed up integrating or the previously suggest approach is incorrect. Thanks in advance.
$\tan \frac{\pi}{3}=\frac{\sin \frac{\pi}{3}}{\cos \frac{\pi}{3}}=\frac{\frac{\sqrt{3}}{2}}{\frac 12}=\sqrt{3}$

$\frac{\tan^7 (\frac{\pi}{3})}{7}=\frac{(\sqrt{3})^7}{7}=\frac{3 \cdot 3 \cdot 3 \cdot \sqrt{3}}{7}=\frac{27 \sqrt{3}}{7}$

Is it the solution ?

5. multiply the integral by $\pi$

$V = \pi \int_a^b (f(x))^2 \, dx$

6. Originally Posted by kalagota
multiply the integral by $\pi$

$V = \pi \int_a^b (f(x))^2 \, dx$
Yes, this is the solution exactly! Thanks for the help.

I hate to be a bother, but I do want to make sure I'm doing these problems right, so I have a similar problem:

"Find the volume of the solid obtained by rotating the region bounded by y = 0, y = e^(5*x), x = 1, and x = ln(3)"

Applying the same idea before would the volume be:

$\pi\int_0^1$e^(10x)dx ?

Or since it's revolved about the y-axis instead of the x-axis would the upper and lower boundaries be different? Thanks again in advance.

7. Originally Posted by auslmar
Yes, this is the solution exactly! Thanks for the help.

I hate to be a bother, but I do want to make sure I'm doing these problems right, so I have a similar problem:

"Find the volume of the solid obtained by rotating the region bounded by y = 0, y = e^(5*x), x = 1, and x = ln(3)"

Applying the same idea before would the volume be:

$\pi\int_0^1$e^(10x)dx ?

Or since it's revolved about the y-axis instead of the x-axis would the upper and lower boundaries be different? Thanks again in advance.
The part is bold is very vague, you forgot to mention what axis your rotating about. Assuming it is the x-axis and I have read the problem correctly the volume is given by $\pi \int_{1}^{\ln 3} e^{10 x} \ dx$.

About the y axis it is a bit more complicated if your sure it's the y-axis I'll post some details on how to do it.

Bobak

8. Originally Posted by bobak
The part is bold is very vague, you forgot to mention what axis your rotating about. Assuming it is the x-axis and I have read the problem correctly the volume is given by $\pi \int_{1}^{\ln 3} e^{10 x} \ dx$.

About the y axis it is a bit more complicated if your sure it's the y-axis I'll post some details on how to do it.

Bobak
Ah yes, I do apologize for not explicitly stating that it was about the y-axis. I greatly appreciate your help.

9. Originally Posted by auslmar
Ah yes, I do apologize for not explicitly stating that it was about the y-axis. I greatly appreciate your help.
ok no forgetting pi this time "::grumbles::"

Ok the formula is

${\color{red}{2\pi}}\int_a^{b}f(x)h(x)$

Where h(x) is the distance from any point in your shaded region that you are attempting to find the volume of to your axis of revolution..in this case it is $h(x)=x$

I believe

10. Yes! Thanks to everyone for the help. The integral is exactly:

$2\pi\int_1^{ln3}xe^{5x}dx$

11. Originally Posted by auslmar
Yes! Thanks to everyone for the help. The integral is exactly:

$2\pi\int_1^{ln3}xe^{5x}dx$
Do you need help evaluating it?

12. Originally Posted by Mathstud28
Do you need help evaluating it?
Oh no, I'm sorry for not mentioning it. I used integration by parts. Thanks again for the help.