Integrating $f(x,y,z)$ within the region of a cylinder and a hyperboloid in cartesian coordinates

Question

I'm confident with solving this question in cylindrical coordinates (for which there are few answers) but I'm doubtful about cartesian coordinates and can't seem to figure out the limits. $z$ needs to be the outer integral in the below question:

Right now, I've figured that:

1. $z$ must vary from $-1$ to $1$ (since it describes the radius of the cylinder).

2. If we make $x$ the inner integral and dependent on $y$ we could say that it varies from $-2y^2$ to $(1-2y^2)$ (obtained by substituting the cylinder equation and adding in $a=0$ & $a=1$

3. I'm not sure how to evaluate the middle integral and how exactly there are 2 regions over which we must integrate.

Answer 1

Hyperboloid $z = \sqrt{a^2+x^2+y^2} \,$ where $a \in (0,1)$ is upper portion of the hyperboloid (above $XY$ plane). Please note $z = -\sqrt{a^2+x^2+y^2}$ gives you a similar region below $XY$ plane.

Given $a \lt 1,$ it does intersect with cylinder $y^2 + z^2 = 1$.

If you visualize the region is bound below and on sides by hyperboloid and bound above by the cylinder.

In cartesian coordinates,

Bounds of $z$ is $ \,\sqrt{a^2+x^2+y^2} \leq z \leq \sqrt{1-y^2} \,$ (bound below by hyperboloid and above by the cylinder)

To find bounds of $x,y$ as you wrote, $1 - a^2 = x^2 + 2y^2 \implies y = \pm \sqrt{\frac{1 - a^2 - x^2}{2}}$

Bounds of $y$, $-\sqrt{\frac{1 - a^2 - x^2}{2}} \leq y \leq \sqrt{\frac{1 - a^2 - x^2}{2}}$

From above equation $1 - a^2 = x^2 + 2y^2$, the lower and upper bounds of $x$ (occurs when $y = 0$) is $\pm \sqrt{1-a^2}$.

Bounds of $x, \, \, -\sqrt{1-a^2} \leq x \leq \sqrt{1-a^2}$

If you need to integrate over both regions above $XY$ plane and below $XY$ plane, just multiply the integral by $2$.

Answer 2

The best way solve problems like this is to try and visualize the solid $V$ in your question. Notice the surface $z=\sqrt{a^2+x^2+y^2}$ is precisely surface formed by rotating the hyperbola $z=\sqrt{a^2+x^2}$ in the $xz-$plane around the $z-$ axis. On the other hand, the cylinder $y^2+z^2=1$ is the surface formed by taking a circle of radius $1$ in the $xz-$plane and translating it across the $x-$axis. You should first notice that $V$ is bounded below by the hyperbloid and above by the cylinder. Moreover, projection of the intersection of these two surfaces onto the $xy-$plane is the ellipse $x^2+2y^2=1-a^2$. So, one way to express this integral would be as $$\int_{-\sqrt{1-a^2}}^\sqrt{1-a^2} \int_{-\sqrt{\frac{1-a^2-x^2}{2}}}^\sqrt{\frac{1-a^2-x^2}{2}} \int_{\sqrt{a^2+x^2+y^2}}^\sqrt{1-y^2}f(x,y,z)dzdydx$$ Try to envision the solid $V$ to see how we can also express this integral as the sum $A+B$ where $$A=\int_a^{\sqrt{\frac{a^2+1}{2}}} \int_{-\sqrt{z^2-a^2}}^{\sqrt{z^2-a^2}} \int_{-\sqrt{z^2-y^2-a^2}}^{\sqrt{z^2-y^2-a^2}}f(x,y,z)dxdydz$$ $$B=\int_{\sqrt{\frac{a^2+1}{2}}}^1 \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} \int_{-\sqrt{z^2-y^2-a^2}}^{\sqrt{z^2-y^2-a^2}}f(x,y,z)dxdydz$$ or even better as $$\int_{-\sqrt{\frac{1-a^2}{2}}}^{\sqrt{\frac{1-a^2}{2}}} \int_{\sqrt{y^2+a^2}}^{\sqrt{1-y^2}} \int_{-\sqrt{z^2-y^2-a^2}}^{\sqrt{z^2-y^2-a^2}}f(x,y,z)dxdzdy$$