# Evaluate $\int_0^1\int_p^1 \frac {x^3}{\sqrt{1-y^6}} dydx$

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?

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

$$\frac{1}{12} \int_0^1 \frac{du}{\sqrt{1-u^2}} = \frac{\pi}{24}$$

• Ah! Ok I need to draw the sketch. But how do I know if my sketch makes any sense at all? I am not very good at graphing. Any tips? Nov 10, 2013 at 12:29
• @Artemisia: you can draw $y=x^2$, right? And you can look sideways...right? At least these were explicit prerequisites for Calc III at my university. That and familiarity with the number $7$. Nov 10, 2013 at 12:31
• @Artemisia: usually for problems like this, the sketches are no more complicated than this. But in general, just make sure that the region is convex and makes sense sideways. Nov 10, 2013 at 12:50
• @GerryMyerson: amazon.com/Penguin-Book-Curious-Interesting-Numbers/dp/… Sep 10, 2015 at 23:58
• Thanks, Ron. I also found this: imdb.com/title/tt0054047 Sep 10, 2015 at 23:59