How to find a volume of an object enclosed with planes without any projection?

Question

How to find a volume of an object enclosed with planes:

$$x^2+z^2=4,$$ $$x+y=2,$$ $$x+y=-2,$$ $$x-y=2,$$ $$x-y=-2$$

without any projection?

When I project this object I know it is a truncated cylinder. So I know I can devide that into 4 (or 8) same parts, count the volume of one part and multiply the result by number of parts to get the volume.

Like this: $$\int_{0}^{2} \int_{0}^{\sqrt{4-x^2}} \int_{-2+x}^{2-x} 1 \,dy\,dz\,dx. $$

This is $\frac{1}{4}$ of the object, so if I multiply it by $4$, I get the volume.

My question is how do I count the volume without making the projection? I think it should be possible but I just don't know how to find the intervals of integration.

Thanks.

Answer 1

Perhaps this will help you visualize the volume:

Answer 2

You can try a change of variables, if only to make the limits easier to identify.

$$\begin{cases} u=x+y\\ v=x-y\\ w=z \end{cases}\iff \begin{cases} x=\frac{u+v}2\\[1ex] y=\frac{u-v}2\\[1ex] z=w \end{cases}$$

Then

$$x^2+z^2=\left(\frac{u+v}2\right)^2+w^2=4\implies w=\pm\frac12\sqrt{16-(u+v)^2}$$

and the Jacobian determinant for this transformation is

$$|\det J|=\left|\det\begin{bmatrix}\frac12&\frac12&0\\\frac12&-\frac12&0\\0&0&1\end{bmatrix}\right|=\left|-\frac12\right|=\frac12$$

The solid (call it $E$) is given by the set

$$E=\left\{(u,v,w)\mid u\in[-2,2],v\in[-2,2],w\in\left[-\frac12\sqrt{16-(u+v)^2},\frac12\sqrt{16-(u+v)^2}\right]\right\}$$

and its volume is then

$$\iiint_E\mathrm dx\,\mathrm dy\,\mathrm dz=\frac12\iiint_E\mathrm du\,\mathrm dv\,\mathrm dw=\frac12\int_{-2}^2\int_{-2}^2\int_{-\frac12\sqrt{16-(u+v)^2}}^{\frac12\sqrt{16-(u+v)^2}}\mathrm dw\,\mathrm dv\,\mathrm du$$

The remaining integrals are doable, if a bit tedious. You would end up with a volume of

$$16\pi-\frac{64}3$$