# Math Help - [SOLVED] Trig integral w/ volume of rotation

1. ## [SOLVED] Trig integral w/ volume of rotation

I know I have my integral correct:

$V = 41\pi \int\limits^{\pi}_{\frac{\pi}{2}} sin^2x dx$

So then I did this:
$
41\pi \int\limits^{\pi}_{\frac{\pi}{2}} \frac{1}{2} (1-cos (2x))$

$
= \frac{41\pi}{2} \int\limits^{\pi}_{\frac{\pi}{2}} 1-cos (2x)$

$
= \frac{41\pi}{2} \int\limits^{\pi}_{\frac{\pi}{2}} 1 - cos (u) du$

$= \frac{41\pi}{4} \cdot (x - \frac{sin2x}{2}) \|^{\pi}_{\frac{\pi}{2}}$

After I evaluate I get $\frac{41\pi^2}{4}$ which is the wrong answer.
I know this is a lot of work, but can someone work this and just point out where my error is? Don't need all the steps, unless I screwed something up.
Thanks!!
Molly

2. You didn't change the integration limits when changed the variable. By the way, that's the correct value for the integral.

Could you post the "correct answer"?

3. But I put 2x back in for "u" and then evaluated at the old limits. Should I not have done that?

This is from that SOLVED problem you helped me with before. Guess I didn't really solve it

Wish I could post the correct answer, but it is in webassign, so I cant get it. I found one in the book that was almost identical:

Find Vol of solid for area btw. curves: y = sinx y=0 with same limits

the book answer for that one is $\frac{\pi^2}{4}$ which is almost exactly what my answer is.

4. The integral: $\int_\frac{\pi}{2}^\pi\sin^2 xdx =\frac{1}{2}\int_\frac{\pi}{2}^\pi 1-\cos 2xdx=\frac{1}{4}\int_{\pi}^{2\pi} 1-\cos tdt=
\left.\frac{1}{4}\left(t-\sin t\right)\right|^{2\pi}_\pi=\frac{\pi}{4}$
. The problem was that you keep the $\frac{1}{4}$ when it should be $\frac{1}{2}$ if you were reversing the change.

It was $(41\sin x)^2$, so the answer should be $41^2\frac{\pi^2}{4}$.

5. OMG! Thank you SOOOOO much! Now I can go to bed! Thanks for all your help tonight!!
Have a great week!

6. There's even a quick way of calculating that integral.

Having $\int_{\frac{\pi }{2}}^{\pi }{{{\sin }^{2}}x\,dx},$ then the substitution $u=\pi-x$ shows that the integral equals $\int_{0}^{\frac{\pi }{2}}{{{\sin }^{2}}x\,dx}=\int_{0}^{\frac{\pi }{2}}{{{\cos }^{2}}x\,dx}$ since $\int_{0}^{\frac{\pi }{2}}{f(\sin x)\,dx}=\int_{0}^{\frac{\pi }{2}}{f(\cos x)\,dx}.$

Finally, $\int_{0}^{\frac{\pi }{2}}{{{\sin }^{2}}x\,dx}+\int_{0}^{\frac{\pi }{2}}{{{\cos }^{2}}x\,dx}=\frac{\pi }{2}$ which implies that both integrals achieve to the same value, $\frac\pi4,$ and the rest follows.