How can I find the volume enclosed by the following graphs?

Question

How do I find the volume bound by planes $x=0$, $y=0$, and the surfaces $x=-4y^2+3$, $z=x^3y$?

I think that I probably need to do:

$$\iint_D (x^3y) \, dx\,dy $$

But I'm not sure what to do for the limits. I think that for the $x$ I should use $0$ and $-4y^2+3$, but I don't know what to do about the $y$.

EDIT: I tried using 0 and 3 for the limits for $x$, and $0$ and $\frac{\sqrt{3-x}}{2}$ as limits for the $y$, but I got half of the answer that I was supposed to get ($\frac{243}{80}$).

Answer 1

Note $y^2=-\frac 14(x-3)$ (See it) thus $$I=2\int_{0}^{3}\int_{0}^{\frac{\sqrt{3-x}}{2}}x^3\,y\,dydx=\frac 14\int_{0}^{3}(3x^3-x^4)dx$$

Answer 2

Alternatively to Behrouz's answer, the solid lies over the plane region $$ D = \left\{(x,y)\mid - \sqrt{3}/2 \leq y \leq \sqrt{3}/2,\ 0 \leq x \leq -4y^2+3\right\} $$ It looks like this:

So the volume is $$ \int_{-\sqrt{3}/2}^{\sqrt{3}/2} \int_0^{3-4y^2} x^3y\,dx\,dy $$