Using triple integrals and cartesian coordinates, find the volume of the solid bounded by $$z=3x^2+3y^2-7 $$ and $$z=9-x^2-y^2$$
My take I have equated the two solids and found the intersection. My first question is, what is exactly intersecting? I am having trouble picturing it. contuniung on The intersection I found is $$x^2+y^2=4$$ and hence the parameters are $$ -2\le x \le 2$$ $$-\sqrt{4-x^2}\le y \le \sqrt{4-x^2}$$ $$3x^2+3y^2-7 \le z \le 9-x^2-y^2 \left( 1 - \frac{y}{b} -\frac{x}{a} \right) $$ and now I am left with this hectic integral $$ \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{x^2+3y^2-7}^{9-x^2-y^2} 1 dzdydx$$ Is there any easier way to solve this integral? Maybe by changing the parameters but I am not sure how?
Use cylindrical coordinates.
$$x=r\cos\theta$$ $$y=r\sin\theta$$ $$z=z$$
$$z=3x^2+3y^2-7\implies z=3r^2-7$$ $$z=9-x^2-y^2\implies z=9-r^2$$
The intersection is now $r=2$.
The volume integral is therefore as follows:
$$V=\int_0^{2\pi}\int_0^{2}\int_{3r^2-7}^{9-r^2}rdz\space{dr}\space{d\theta}$$
This is a much easier integral to calculate.