I'm trying to solve what I thought was a fairly simple integration problem where I need to find the volume of something, but I've gotten stuck.
I'm using the "disk/washer" method, and the integral I need to solve is this:
$$ \int_0^{\pi \over4} \pi\left(\left(2-\sin(x)\right)^2 - \left(2-\cos(x)\right)^2 \right)dx $$
I've checked with integral-calculator.com, and I'm getting the correct anti-derivative, but when I try to evaluate over $[0,{\pi\over 4}]$ I'm not getting the right answer.
On integral-calculator, they say the anti-derivative is:
$$-2\pi\left(\sin(2x)-4\left(\sin(x) + \cos(x) \right) \right)+C$$
Which is exactly what I got when I tried to solve this. From here, I did the usual substitution of $\pi \over 4$ etc... but the answer I'm getting is negative and obviously wrong.
On integral-calculator, they go from that anti-derivative, to this:
$$-\pi\left( {5\cdot2^{3\over2} -16 }\over\sqrt{2} \right)$$
I'm at a loss here. This answer is correct, but I can't figure out how to get from the anti-derivative, to this "simplification", and integral-calculator doesn't have any elaboration on how they got there.
Any assistance is greatly appreciated as I would really like to understand what is going on here. Thanks.
First evaluate at $\pi/4$: $$-2\pi (\sin(\pi /2)-4\sin(\pi/4)-4\cos(\pi/4))=2\pi(-1+4 \sqrt 2)$$ Then evaluate at $0$: $$-2\pi (\sin(0)-4\sin(0)-4\cos(0))=8\pi$$ Then subtract: $$2\pi (-1+4\sqrt 2)-8\pi=2\pi(-5+4\sqrt 2)=-\pi\bigg(\frac{5\cdot 2^{3/2}-16}{\sqrt 2}\bigg)$$