TITLE: Evaluate $\int_0^1\int_p^1 \frac {x^3}{\sqrt{1-y^6}} dydx$
QUESTION [6 upvotes]: I have been working on this sum for a while. The question asks to evaluate the double integral.
$$\int_0^1\int_p^1 \frac {x^3}{\sqrt{1-y^6}} dydx$$
where $p$ is equal to $x^2$. I know that I have to solve the $y$ integral first and then the $x$. But I don't know how to solve the root integral. Applying the formula $$\int \frac{1}{\sqrt{1-t^2}}dt$$ where $t=y^3$ isn't working. Any ideas as to how I must proceed with the integral? Once I get the integral, I must substitute the limits and then the integral would be in terms of $x$ and I must integrate it. Am I correct?

REPLY [15 votes]: Change the order of integration.  You can do this by drawing the region of integration and seeing that the integral is just
$$\int_0^1 \frac{dy}{\sqrt{1-y^6}} \, \int_0^{\sqrt{y}} dx \, x^3$$
which is
$$\frac14 \int_0^1 dy \frac{y^2}{\sqrt{1-y^6}}$$
or
$$\frac{1}{12} \int_0^1 \frac{du}{\sqrt{1-u^2}} = \frac{\pi}{24}$$