Find the volume of a solid under the surface $z=x^{2} -y^{2}$ and limited by the planes: $xy$, $x=2$ and $x=4$.
Can you please tell me whether I'm approaching this problem correctly?
As the region is limited by the $xy$ plane, $z\geq0$. So the volume asked will be $\iint\limits_Rf(x,y)dxdy$ where $R$ is the base of the surface (the projection on $xy$).
We define $R$ more accurately as a y-simple region. So $2\le x \le 4$. The bounds of the projection on $xy$ will be those where $0=x^{2} -y^{2} \implies -x\le y \le x$.
The last step is to set the limits for $R$ in the double integral and integrate.$\int\limits_2^4\int\limits_{-x}^xx^{2}-y^{2}dydx=80$
Best Answer
To formalize my comments, I will put it in answer form. First off, your solution is correct. Just to add some clarification, the $xy$ plane is equivalently represented as $z=0$. The triple integral, $\int \int \int_0^{x^2-y^2} dzdydx$ reduces to a double integral that you described above. Next, we take the projection of the surface on the $xy$ plane, by finding the intersection of $z=x^2-y^2$ and $z=0$ which is simply $x^2=y^2$. Following that we draw the projection and deduce the following,
$$V=\int_{2}^{4} \int_{-x}^{x} \int_0^{x^2-y^2} dzdydx$$
The projection should look something like this (the red line is the ray that is drawn to determine the bounds)