Revolving trig function around y-axis

[metalclay]

Homework Statement

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

Homework Equations

tan^2=sec^2 - 1

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!

 

[Ray Vickson]

Homework Statement

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

Homework Equations

tan^2=sec^2 - 1

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!

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".

 

[metalclay]

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".

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

http://imgur.com/a/SaVJF

 

[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

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!

 

[SammyS]

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!

It looks OK.

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

 

[metalclay]

It looks OK.

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

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

Ha, thanks guys!