<s>[INST] How to integrate with a function in the limit $\int ^a _0 (\int_0 ^ \sqrt {a^2 - x^2} (a^2 - x^2 - y^2)dy)dx = 303$
I'm trying to solve the above question where the integral has a function as the limit. 
Seeing the similarity between the limit and the variable integration, I substituted the $\sqrt {a^2 - x^2}$ with $u$. 
Hence I end up with:
$\int ^a _0 (\int_0 ^ u (u^2 - y^2)dy)dx = 303$
After that, I'm not too sure how to proceed as there isn't much to go on as far as integrating $y$ with respect to $u$ is concerned. 
EDIT:
Building onto the advice given:
$\int ^a _0 (\int_0 ^u (u^2 - y^2)dy)dx = 303$
$\int ^a _0 [u^2y - \cfrac{y^3}{3}]^u _0 dx = 303$
$\int ^a _0 [u^3 - \cfrac{u^3}{3}] dx = 303$
$[u^3x - \cfrac{u^3x}{3}]^a_0$ = 303
$((a^2 - a^2)^{1.5}a - \cfrac{(a^2 - a^2)^{1.5}a}{3} = 303$
EDIT1:
$\int ^a _0 (\int_0 ^\sqrt{a^2 - x^2} (a^2 - x^2 - y^2)dy)dx = 303$
$\int ^a _0 [a^2y - x^2y - \cfrac{y^3}{3}]^\sqrt{a^2 - x^2}_0 dx = 303$
$\int ^a _0 [a^2\sqrt{a^2 - x^2} - x^2\sqrt{a^2 - x^2} - \cfrac{((a^2 - x^2)^{1.5})}{3}dx = 303$
$\int ^a _0 [(a^2 - x^2)\sqrt{a^2 - x^2} - \cfrac{((a^2 - x^2)^{1.5})}{3}dx = 303$
$\int ^a _0 [(a^2 - x^2)^{1.5} - \cfrac{((a^2 - x^2)^{1.5})}{3}dx = 303$
$\int ^a _0 [\cfrac{2}{3}(a^2 - x^2)^{1.5}dx = 303$
 [/INST] As a general rule, whenever possible, try to draw a picture of the region over which you are integrating. In this case, the limits of integration describe a quarter circle in the first quadrant of radius $a$, centered at the origin. This suggests a change of variable to polar coordinates, which turns the integral into
$$\int_0^{\pi/2}\int_0^a(a^2-r^2)r\,dr\,d\theta={\pi\over2}\int_0^a(a^2-r^2)r\,dr={\pi\over2}a^4\int_0^1(1-u^2)u\,du={\pi\over2}a^4\left({1\over2}-{1\over4}\right)={\pi\over8}a^4$$
Remark: It strikes me as a little odd that the problem would set the integral equal to $303$. I would have expected $32\pi$ instead (for which the solution would simply be $a=4$). But so be it.
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