<s>[INST] Find $\iint_{D} y^{3}\,dx\,dy$ $$\iint_{D} y^{3}\,dx\,dy$$ where $D$ is the domain between $x^2 + y^2 = 6$ circle and the parabola $y=x^2$
Edit: Also, I got the intersection of the curves $(\sqrt2,2)$ and $(\sqrt-3, -3)$
I draw the curves but I got confused at the bounds of the integrals.
 [/INST] Double-check your work finding the points of intersection, as they should be $\left(-\sqrt{2},2\right)$ and $\left(\sqrt{2},2\right)$. These give $x=-\sqrt{2}$ and $x=\sqrt{2}$ as your lower and upper bounds for $x$, respectively.

Looking at the plot provided, it is clear that the lower and upper bounds for $y$ will be the parabola $y=x^2$ and the upper branch of the circle $x^2+y^2=6$ ($y=\sqrt{6-x^2}$), respectively. Therefore,
\begin{align*}
\iint_{D}y^3 dA &= \int_{-\sqrt{2}}^{\sqrt{2}}\left(\int_{x^2}^{\sqrt{6-x^2}}y^3 dy\right)dx\\
&= \int_{-\sqrt{2}}^{\sqrt{2}}\left(\frac{\left(\sqrt{6-x^2}\right)^4}{4}-\frac{\left(x^2\right)^4}{4}\right)dx\\
&= \frac{1}{4}\int_{-\sqrt{2}}^{\sqrt{2}}\left((6-x^2)^2-x^8\right)dx\\
&= \frac{1}{2}\int_{0}^{\sqrt{2}}(6-12x^2+x^4-x^8)dx\\
&= \frac{1}{2}\left(6\sqrt{2}-4\left(\sqrt{2}\right)^3+\frac{\left(\sqrt{2}\right)^5}{5}-\frac{\left(\sqrt{2}\right)^9}{9}\right)\\
&= \frac{608\sqrt{2}}{45}
\end{align*}
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