I am trying to calculate the following integral
$$V = \pi \int_{-16}^{-7} \left( \sqrt{256 - y^2} \right)^2 \, \mathrm{d}y $$
I get the following result: $1053 \pi$.
But when I use the usual transformation ($x=r\cos(\theta)$, $y=r\sin(\theta)$), I get the following integral
And it results $760.43$. I should get the same results with both integrals, what am I doing wrong? I have done the transformation like 10 times :(
To integrate, we split up the terms and use a $u$-substitution ($u = sin(\theta)$) in the second term:
$I(\theta)={\displaystyle\int}{\pi}\left(16^3\cos\left({\theta}\right)-16^3\cos\left({\theta}\right)\sin^2\left({\theta}\right)\right)\,\mathrm{d}{\theta}$=${\displaystyle\int}{\pi}\left(4096\cos\left({\theta}\right)-4096\cos\left({\theta}\right)\sin^2\left({\theta}\right)\right)\,\mathrm{d}{\theta}= {\displaystyle\int}4096{\pi}\cos\left({\theta}\right)\,\mathrm{d}{\theta} -{\displaystyle\int}4096\pi\cos\left({\theta}\right)\sin^2\left({\theta}\right)\,\mathrm{d}{\theta} =4096{\pi}\sin\left({\theta}\right)-\dfrac{4096{\pi}\sin^3\left({\theta}\right)}{3} +C$
Now plug in your bounds for $-\frac{\pi}{2}\leq \theta \leq -\arcsin(\frac{7}{16})$ to get the answer:
$I(-\arcsin(\frac{7}{16}))-I(-\frac{\pi}{2}) = 4096\pi\{-\frac{7}{16}+1\}-\frac{4096\pi}{3}\{-\frac{7^3}{16^3}+1\}=4096\pi(\frac{9}{16})-4096\frac{\pi}{3}(\frac{3753}{4096})=16^2\cdot9\pi-1251\pi=1053\pi$