Finding integration bounds for the volume between ellipsoid and plane

Question

Let $S$ be the region between $x+z=1$, and ellipsoid $x^{2}+2y^{2}+z^{2}=1$ ($S$ is above $x+z=1$). Find the volume of $S$.

I'm having trouble to find the bound for $x,y$, and $z$. At first I tried to identify the bound of $z$ as follows:

$1-x\le z\le \sqrt{1-x^{2}-2y^{2}}\quad$ and for $x$ is $0\le x\le1\quad$.

But I struggled to find the bound for $y$.

My idea for this is to find the ellipse equation on $x+z=1$ plane to substitute $z=1-x$ to the ellipsoid equation to get $(2x-1)^{2}+(2y)^{2}=1$.

So I get: $-\sqrt{-(x-\frac{1}{2})^{2}+\frac{1}{4}}\le y\le\sqrt{-(x-\frac{1}{2})^{2}+\frac{1}{4}}$

Is this right or am I doing it wrong?

Answer 1

This is how I would find the volume bounded by the ellipsoid $x^2 + 2y^2 + z^2 = 1$ , and above the plane $x + z = 1 $.

First transform the ellipsoid into a sphere, by defining the transformation:

$ x' = x , y' = \sqrt{2} y , z' = z $.

That is,

$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 && 0 && 0 \\ 0 && \dfrac{1}{\sqrt{2}} && 0 \\ 0 && 0 && 1 \end{bmatrix} $ $ \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix} $

Then, in the transformed variables, we have

$ x'^2 + y'^2 + z'^2 = 1$ and $ x' + z' = 1$

Now we're going to work with these transformed variables. The plane $x'+z' = 1$ cuts the sphere $x'^2 + y'^2 + z'^2 = 1$ in a circle. To find its center and radius, find the distance between the origin and the plane. This distance is

$ d = \dfrac{1}{\sqrt{2}} $

Therefore the radius is

$ R = \sqrt{ 1 - \left( \dfrac{1}{2} \right)^2 } = \dfrac{\sqrt{3}}{2} $

And the center is

$ C = d \dfrac{(1, 0, 1)}{\sqrt{2}} = \dfrac{1}{2} (1, 0, 1) $

The unit normal to the plane is $N = \dfrac{(1, 0, 1)}{\sqrt{2}} $

A plane parallel to the cutting but shifted a distance $t$ in the direction of $N$ will have the equation $ N \cdot ( r' - (C + t N) ) = 0 $ where $r' = (x', y', z') $.

The range for $t$ is from $0$ to $(1 - d) = 1 - \dfrac{1}{\sqrt{2}} $. At the upper limit the shifted plane becomes tangent to the sphere.

The shifted plane cuts the sphere in a circle whose radius is $r(t)$. To find $r(t)$, find the distance between the origin and the shifted plane. This is

$ s(t) = | N \cdot (C + t N) | = \dfrac{1}{\sqrt{2}} + t $

So the square of the radius of the circle cut by the shifted plane is

$ r^2(t) = 1 - s^2(t) = \dfrac{1}{2} - \sqrt{2} t - t^2 $

Therefore the volume in the transformed variables is

$ V' = \displaystyle \int_{0}^{1 - \frac{1}{\sqrt{2}}} \pi r^2 dt = \displaystyle \int_{0}^{1 - \frac{1}{\sqrt{2}}} \pi (\dfrac{1}{2} - \sqrt{2} t - t^2) dt \\ = \pi \bigg[ \dfrac{1}{2} t - \dfrac{t^2}{\sqrt{2}} - \dfrac{1}{3} t^3 \bigg] \bigg|_{0}^{1 - \frac{1}{\sqrt{2}}} = \pi \left( \dfrac{2}{3} - \dfrac{5 \sqrt{2}}{12} \right) $

Finally, to find the actual volume in the original variables, we have to multiply $V'$ by the determinant of the transformation matrix. This determinant is $\dfrac{1}{\sqrt{2}}$. Therefore, the volume is

$ V = \pi \left( \dfrac{\sqrt{2}}{3} - \dfrac{5}{12} \right) $

Answer 2

The projection of the intersection curve of surface $x+z=1$ with the surface $x^2+2y^2+z^2=1$ on $xy$-plane is the circle $x^2+y^2=x$ that you found. In polar coordinates this circle is: $$r^2=r\cos\theta\implies \bbox[5px,border:2px solid black]{r=\cos\theta}\tag1$$ The solid $S$ we consider is bounded from above by $$x^2+2y^2+z^2=1\implies\bbox[5px,border:2px solid black]{z=\sqrt{1-(1+\sin^2\theta)r^2}}\tag2 $$ and from below $$x+z=1\implies\bbox[5px,border:2px solid black]{z=1-r\cos\theta}\tag3$$ The projection of the solid on $xy$-plane is the region bounded by the circle $(1)$: $$\bbox[10px,border:2px solid black]{ \begin{align}-\frac\pi 2\leq\theta\leq\frac\pi 2\\ 0\leq r\leq\cos\theta \end{align}}\tag4$$ From $(2),(3)$ and $(4)$, the volume integral of $S$ is $$V=\int_{-\frac\pi 2}^{\frac\pi 2}\int_0^{\cos\theta}\left( \sqrt{1-(1+\sin^2\theta)r^2}\\-(1-r\cos\theta)\right) rdrd\theta.$$