How to integrate $\int_{y=0}^a (y^2 \sqrt{a^2-y^2}) dy$ I am having trouble integrating $\int_{y=0}^a (y^2 \sqrt{a^2-y^2}) dy$ . If possible, could someone show me the detailed steps for how to get the answer ${a^4 \pi \over 16}$ ? 
(I think you have to use trig substitution, but how would you do that?)
 A: Hint: If $y=a\sin t$ for $t$ in $\left(0,\frac\pi2\right)$, then $\mathrm dy=a\cos t\mathrm dt$ and $\sqrt{a^2-y^2}=a\cos t$.
A: Use the substitution $y=a\sin\theta \Rightarrow dy=a\cos\theta\,d\theta$ to get:
$$a^4\int_0^{\pi/2} \sin^2\theta \cos^2\theta\,d\theta=\frac{a^4}{4}\int_0^{\pi/2}\sin^2(2\theta)\,d\theta=\frac{a^4}{8}\int_0^{\pi/2}\left(1-\cos(4\theta)\right)\,d\theta$$
$$=\boxed{\dfrac{a^4\pi}{16}}$$
A: Square roots of the form $\sqrt{a^2-x^2}$ are hard to integrate without a change of variable so we would like to do some change of variable that makes this simpler. Enter trig functions. We know that $\sin^2xt+\cos^2t=1$ so $\sin^2t = 1-\cos^2t$ or $\cos^2t=1-\sin^2t$. Note how similar this looks to the expression inside our square root.
This suggests that maybe we should make a change of variable of $y = a\sin t$ so that $y^2 = a^2\sin^2t$ and our square root becomes
$$\sqrt{a^2-a^2\sin^2t} = \sqrt{a^2(1-\sin^2t)} = |a|\sqrt{1-\sin^2t} = |a|\sqrt{\cos^2t} = |a||\cos t|.$$
However we can actually say something stronger. We can omit the absolute value by assuming $a>0$ which is not a big assumption since we are integrating an even function. Further if $y=0$, then $t = 0$ and if $y = a$, $t = \frac{\pi}{2}$ and for $t$ in this range, $\cos$ is positive so the absolute value around $\cos$ does not play a role. Hence our integrand becomes
$$y^2\sqrt{a^2-y^2} \Longrightarrow a^2\sin^2t\cos t.$$
Do not forget to adjust your differential though since chain rule is now a factor here.
A: As others have pointed out, this problem is done by trigonometric substitution. If you can just look at it and see $y=a \sin(\theta)$ then that's great; I was never able to do that. Instead, I learned to identify the substitution by drawing a triangle. Draw a right triangle; pick one of the acute angles to be $\theta$. Ultimately we want to make the triangle so that $\sqrt{a^2-y^2}$ and $y^2$ are "simple" trigonometric functions of $\theta$.
To do this, make the hypotenuse be length $a$ and the side opposite $\theta$ be length $y$. Then the adjacent side is length $\sqrt{a^2-y^2}$. So now we can see that $\sin(\theta)=y/a$ and $\cos(\theta)=\sqrt{a^2-y^2}/a$. This lets us write $y^2=a^2 \sin(\theta)^2$, $dy=a \cos(\theta) d\theta$, and $\sqrt{a^2-y^2}=a\cos(\theta)$. Then $\theta(y)=\sin^{-1}(y/a)$ so $\theta(a)=\pi/2$ and $\theta(0)=0$. So
$$\int_0^a y^2 \sqrt{a^2-y^2} dy = \int_0^{\pi/2} a^4 \sin(\theta)^2 \cos(\theta)^2 d \theta$$
You can do this as others have already shown. It is instructive to make the adjacent side be $y$ instead, and observe that it does not change the answer.
