Need guidance simplifying $\iint (2-x^4-y^4)\sqrt{6-x^2} \,d\mathbf A$

Question

we have $$\iint (2-x^4-y^4)\sqrt{6-x^2} \, d\mathbf A \qquad(-1\le x \le 1), \quad(-1\le y \le 1)$$

i have attempted converting to polar coordinates and got $$\iint (2-(r\cos(\theta))^4)-(r\sin(\theta))^4)\sqrt{6-(r\cos(\theta))^2}(r) \, d\mathbf A$$ However it doesnt simplify nicely and i am having a hard time seeing where to go from here. Can someone point me in the right direction?

thanks

Answer 1

COMMENT.- Without region $A$ (you can do this), integrate first respect to $y$ so you have $$y\int2x^4\sqrt{6-x^2}dx- \frac {y^5}{5}\int\sqrt{6-x^2}dx+ax+b$$ and $$y\int2x^4\sqrt{6-x^2}dx=\frac{yx}{2}\sqrt{6-x^2}(x^2-3)+9\arcsin\left(\frac{y}{\sqrt6}\right)+c$$ Try to calculate the second integral and evaluation over the region $(-1\le x \le 1)(-1\le y \le 1)$ to finish.