To find volume using triple integration

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Find volume of the solid above XY plane and directly below the portion of the elliptic paraboloid $x^2+\frac{y^2}{4}=z$ which is cutoff by plane $z=9$

Now I came up with

$\int \int x^2 + \frac{y^2}{4} $ dxdy. After using change of variable as $x = 3rcos\theta$ and $y =6rsin\theta $. I got the integral as

$\int_{0}^{2\pi} \int_{0}^{1} 9r^2 \cdot18 \cdot r drd\theta$ = $81 \pi$

Please check if this is correct or not ?

Thanks

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Yup, sounds good.

The Jacobian of $3r\cos(\theta)$ and $6r\sin(\theta)$ is $18r$, so that's good.

And then, $(3r\cos(\theta))^2+\frac{1}{4}(6r\sin(\theta))^2=9r^2$, so that works out too.

Finally, your bounds of integration are valid and so is your answer!