# Homework Help: Revolving trig function around y-axis

1. Sep 5, 2016

### metalclay

1. The problem statement, all variables and given/known data
http://imgur.com/a/emr1n
01 x= 4tan(pi*y/4) , Find the volume of the solid by revolving shaded region about the y-axis

2. Relevant equations
tan^2=sec^2 - 1

3. The attempt at a solution
I ended up with 64-16pi using u-substitution, not sure if right, want a confirmation. I only have one last attempt on mathlab so really need this to be right, would like to know how you got to whatever the answer is, or if there's a way to check the answer, that would help a lot. Wolfram hasn't been of much help :/

Thanks!

Last edited: Sep 5, 2016
2. Sep 5, 2016

### Ray Vickson

What formula did you use for the volume? What was your integral before and after u-substitution? We are forbidden from doing the problem for you, so we cannot show you "...how you got to whatever the answer is".

3. Sep 5, 2016

### metalclay

yeah, sure. It's just that it's kind of long so didn't know if I should post:

http://imgur.com/a/SaVJF

4. Sep 5, 2016

### metalclay

basically the area of a circle by its height dx

substitute the equation f(y) = 4tan (pi*y/4) for r

so r^2 is tan^2 which is (sec^2 - 1)

found the integral by u substituting (pi*y/4) with u

so you can now integrate sec^2 and 1

which gives tan (u) - u

plugged those values in multiplied by the scalars I got from before which were 16*4pi

and after all that, give or take, I got 64-16pi. I'm not sure that's the right answer (it could be) I just only have one more shot on MathLab, and want to get someone who knows what they're talking about to chime in before I submit :/

Thanks!

Last edited: Sep 5, 2016
5. Sep 6, 2016

### SammyS

Staff Emeritus
It looks OK.

I get the same result using a somewhat different set of steps.

6. Sep 6, 2016

### metalclay

bit the bullet and just clicked "final check" and....*drumroll* it's right!

Ha, thanks guys!