Volume Enclosed Between a Surface and a Plane

Question

This is a problem that I came up with that has been driving me crazy. I have not been able to find a solution nor a person who could direct me to one. The problem is as follows; Find the volume of the region enclosed by the surfaces on [$-\pi..\pi, -\pi..\pi]$ (These are meant to describe the x and y value ranges, excuse my lack of knowledge on formal math notation) $$F(x,y)=cos(x)+cos(y)$$ $$G(x,y)=1/2$$ I need to specific, however, that I do not want to find the net volume of this region as 1-These are trigonometric functions with an infinite domain and therefore never reach a truly enclosed area and 2-Selecting a square region as I have will result in an answer of 0. The volume I want to find in specific is the one enclosed by the plane G(x,y) and F(x,y) where F(x,y) lies above the plane. Below are some pictures to illustrate this.

This problem is a bit above my current math knowledge, but I am really itching for a solution/walkthrough on the problem. The biggest obstacle for me is determining how to describe the boundary of the region I need to integrate over. Any help would be much appreciated and I can answer any questions as I know my explanation isn't perfectly clear/formalized.

Answer 1

So it is clear that the region of integration reaches its maximum width when either $x$ or $y$ $=0$. Setting $y=0$, $$F(x,0)=\cos(x)+\cos(0)$$ Then, $$F(x,0)=G(x,y) \implies \cos(x)+\cos(0)=\frac{1}{2}$$ Which has solutions at $x=\pm \frac{2\pi}{3}$. Now let's find the bounds for our $y$ values. So, $$F(x,y)=G(x,y) \implies \cos(x)+\cos(y)=\frac{1}{2}$$ So then $$y=\pm \arccos(\frac{1}{2}-\cos(x)).$$ So our integral is $$\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}\int_{-\arccos(\frac{1}{2}-\cos(x))}^{\arccos(\frac{1}{2}-\cos(x))}(\cos(x)+\cos(y)-\frac{1}{2}) \mathrm{d}y\mathrm{d}x \approx 8.26979$$ Computed here

Answer 2

If I understand correctly, your question is to find the volume bounded in the "hump" (call it $R$) where $F > G$. This will involve integrating a function with respect to $x$ and $y$ over some region $R$ corresponding to the points where $F > G$. That is, $F(x, y) > G(x, y)$ if and only if $(x, y) \in R$.

To do this, we just need to find what the ranges of $x$ and $y$ are that define the region $R$. We can observe that boundary of $R$ is defined by the relation $\cos (x) + \cos (y) = 1/2$. Note that both $\cos (x)$ and $\cos (y)$ are both at least $-1/2$ for the equality to hold, since $\cos$ takes on values in the range $[-1, 1]$. Thus, $x, y \in [-2\pi/3, 2\pi/3]$.

Let us calculate the volume of the region where $x, y \geq 0$. Note that this constitutes a quarter of the hump (let's call this portion $Q$). As with $R$, the region $Q$ is defined by the inequality $\cos (x) + \cos (y) \geq 1/2$, so for any fixed $x$, $\cos (y)$ can range from $1$ to $1/2 - \cos(x)$, so $y$ can range from $0$ to $\arccos(1/2 - \cos x)$. The volume of the region $H$ is given by the integral $$\int_H [F(x, y) - G(x, y)] dx dy = \int_0^{2\pi / 3} \int_0^{\arccos(1/2 - \cos x)} [\cos x + \cos y - 1/2] dy dx$$ To get the area of your desired region $R$, you would just have to multiply the result of this integral by four. Unfortunately, this integral does not seem to have a closed-form solution (the $\arccos(1/2 - \cos x)$ causes quite a few problems), so numerical integration seems to be the way to go.