Volume of a solid enclosed by the cone $4y^2=x(2-z)$ and the planes $x+z=2,z=0$

Question

This question has already been asked long ago, so I didn't know if I should post my answer and I was also wondering if this could be done smoothly with some change of variables:

Compute the volume of a solid enclosed by the cone $4y^2=x(2-z)$ and planes $x+z=2,z=0.$

Source: Berman, task 3585

My attempt:

The solid we're looking at is $$S=\{(x,y,z)\in\Bbb R^3\mid 0\le x\le 2, 0\le z\le 2-x,4y^2\le x(2-z)\}.$$

$$\begin{aligned}&\color{white}=\int_0^2\int_0^{2-x}\int_{-\frac{\sqrt{x(2-z)}}2}^{\frac{\sqrt{x(2-z)}}2}dydzdx\\&=\int_0^2\int_0^{2-x}\sqrt{x(2-z)}dzdx\\&=\int_0^2\sqrt x\int_0^{2-x}\sqrt{2-z}dzdx\\&=\begin{bmatrix}t^2=2-z\implies 2tdt=-dz\end{bmatrix}\\&=\int_0^2\sqrt x\int_{\sqrt 2}^{\sqrt x}-2t^2dt\\&=2\int_0^2\sqrt x\int_{\sqrt x}^{\sqrt 2}t^2dt\\&=\frac23\int_0^2(2\sqrt 2\sqrt x-x^2)dx\\&=\frac23\left(\frac{4\sqrt 2}3x^{3/2}-\frac{x^3}3\right)\Big|_0^2\\&=\frac{16}9\end{aligned}$$

This seems to be a rotated elliptic cone. I think I understand how to obtain the equation of a rotated circular cone the way its done here. I tried to find a transformation and a cone in the standard position that would be mapped to the one in task in order to find some better parametrization, but I failed.

Is there any option, other than Cartesian coordinates, that wouldn't make the integral more complicated?

Answer 1

We can translate the coordinate system along z-axis such that the vertex of the cone is at $(0, 0, 0)$ by using $Z = z-2$.
Or we can use change of variable $~Z = 2 - z$, that takes the lower half of the cone to above Z-plane.

Then the equation of cone is $~4y^2 = x \cdot Z$

Plane $x + z = 2$ is to be written as $x = Z$ and $z = 0$ as $Z = 2$.

In cartesian coordinates, going in the order $dx ~ dy ~ dZ$,

$ \displaystyle \frac{4y^2}{Z} \leq x \leq Z$

At the intersection of plane $x = Z$ and cone $4y^2 = xZ$,

$ \displaystyle y = \pm \frac{Z}{2}$

So the volume is,

$ \displaystyle \int_0^2 \int_{-Z/2}^{Z/2} \int_{4y^2/Z}^{Z} dx ~ dy ~ dZ = \frac{16}{9}$

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On your question about using other coordinate systems without making the integral more complicated, in addition to translating the coordinate system along z-axis, you will also need rotation of xz plane, either $45^\circ$ clockwise or $45^\circ$ anti-clockwise. That leads to axis of the cone aligned to either x-axis or z-axis. Then you can use either the cylindrical coordinates or spherical coordinates by transforming the elliptic cone to a circular cone.