How to solve this double integral?

I can put this equation

$$\int_0^{\sqrt{2}}\int_{x^2}^2 x (y^2+2)^{\frac{1}{4}} \, dy ~dx$$

into Wolfram but I don't see the intermediate steps and don't know how to solve this directly. I tried u-substitution but that doesn't seem to apply. It might be a hypergeometric but I don't know how to tackle it.

PS: This is not homework

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Try interchanging the order of integration. Be careful to find the new limits of integration. –  David Mitra Feb 13 '13 at 14:54
How do you mean? –  user51819 Feb 13 '13 at 15:00
Try Derive 6 by Texas Instruments. It'll display steps in the solution. –  Student Feb 13 '13 at 15:06
He means $\int_0^{\sqrt 2}\left[\int_{x^2}^2 x(y^2+2)^{1/4}dy\right]dx =\int_?^{?}\left[\int_{?}^? x(y^2+2)^{1/4}dx\right]dy$ –  GEdgar Feb 13 '13 at 15:09
I assume your integral is $\int_0^{\sqrt 2}\int_{x^2}^2 x(y^2+2)^{1/4}\,dy\, dx$. Sketch the region of integration. It's a "quarter circleish" shape. Now write the integral so that you're integrating with respect to $x$ first. You'll obtain $\int_0^2\int_0^{\sqrt y} x(y^2+2)^{1/4}\,dx\,dy$. –  David Mitra Feb 13 '13 at 15:09

Perhaps you meant:

$$\int\limits_0^{\sqrt 2}\int\limits_{x^2}^2x\sqrt[4]{y^2+2}dydx=\int\limits_0^2\int\limits_0^{\sqrt y}x\sqrt[4]{y^2+2}dxdy=\frac{1}{2}\int\limits_0^2y\sqrt[4]{y^2+2}\;dy=$$

$$=\frac{1}{4}\int\limits_0^2(2y\,dy)(y^2+2)^{1/4}=\left.\frac{1}{4}\frac{4}{5}(y^2+2)^{5/4}\right|_0^2=\frac{1}{5}\left[6^{5/4}-2^{5/4}\right]$$

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Draw the curve $y=x^2$, and the line $y=2$, and the line $x=\sqrt{2}$.

Our integral has $y$ going from $x^2$ to $2$, and then $x$ going from $0$ to $\sqrt{2}$.

So we are integrating over the part of the first quadrant which is above $y=x^2$, and below $y=2$.

We don't really want to try to find an antiderivative of $(y^2+1)^{1/4}$. In fact, this function does not have an elementary antiderivative. But at least it looks hard.

So integrate first with respect to $x$, from $x=0$ to $x=\sqrt{y}$. Then let $y$ go from $0$ to $2$.

When we integrate $x$, we get $x^2/2$. When we evaluate at $\sqrt{y}$ and $0$, we end up with an expression of shape $ky(y^2+2)^{1/4}$. This integrates easily.

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