"Find the volume of the 3-dimensional region with $𝑥>0, 𝑦>0, 𝑧>0$ given by
$$𝑧^2<𝑥+𝑦<2𝑧$$
So I have tried many approaches.
First I obviously figured out the boundaries of the shape. They should be 4 for both x and y and 2 for z.
Then, I worked out that horizontal cross-sections are triangles with area $\frac{1}{2}xy$ and I assumed that we can take isosceles triangles so area is $\frac{1}{2}x^2$ I then took and integral multiplying area by height $$\int_0^4\frac{1}{2}x^2(\sqrt{x}-\frac{x}{2})dx$$
That was wrong.
I then tried to take vertical cross sections of the necessary area between $z^2$ and 2z curves. I figured that they should have rectangular shapes where height is the z coordinate difference between $z^2$ and 2z, width is the distance from x-axis to y-axis which I estimated to $x\sqrt{2}$ or $y\sqrt{2}$ again using isosceles triangles properties and thickness is dx and then set up and integral $$\int_0^4x\sqrt{2}(\sqrt{x}-\frac{x}{2})dx$$ didn't work either.
I have calculated the volume of the pyramid bounded by 2z curve and got around 9.98.
In general I understand that I should take $\frac{1}{2}xy$ and integrate it from 0 to 2 along dz for z^2 and 2z curves but I have a few variables pop up and I don't know how to do it without multivariable calculus.
Yes, just to say. This is a single variable calculus course yet, so I don't think we can use double or triple integrals.
I have a feeling that I go wrong because I assume that we have to integrate along dx or dy axis whether it might not be the case and I struggle to understand how to integrate along z-axis.
Anyways, if you have any idea regarding this one. Would appreciate some help.
Thanks
Best Answer
Since it is supposed to be solved as a 1-dimensional integration problem, most probably Cavalieri's principle is to be used.
$$ A(z) = \frac 12 \cdot (2z)^2 - \frac 12\cdot (z^2)^2= 2z^2-\frac{z^4}{2}$$
All together
$$V = \int_0^2\left( 2z^2-\frac{z^4}{2}\right)dz=\frac{32}{15}$$