the volume of the solid limited by the surfaces $x^2+z^2=4$ and $y^2+z^2=4$.

Question

I want to find the volume of the solid limited by the surfaces $x^2+z^2=4$ and $y^2+z^2=4$.

I have broken the volume into sixteen pieces, one of which can be represented by the following section:

$0\leq x \leq 2 $

$0\leq y \leq x $

$0\leq z \leq \sqrt{4-y^2} $

Whenever I integrate over this section, I get a volume with $\pi$, whilst the answer I want is $\dfrac{128}{3}$.

I used the integral:

$$\int_{0}^{2}\int_{0}^{x} \sqrt(4-y^2) dydx$$

I solved this with trig substitution, which left me with an inverse sin which went to $\dfrac{\pi}{2}$.

Yes, it is. I should have said.
The problem has been reduced to one half of the volume in the 1st octant. — Ian, Jul 17 '16 at 22:48
I've edited your latex slightly (last line); let me know if that broke anything. Why don;t you show us your work that led to the answer "$\pi$"? That way, we can perhaps help you learn how to debug such things. — John Hughes, Jul 17 '16 at 22:50
If you either want $z$ from $0$ to $\sqrt {4-x^2}$, or $y$ from $x$ to $2.$ As these are two cylinders intersecting, I would also consider cylindrical coordinates. — Doug M, Jul 17 '16 at 22:58
This sort of shapes is known as Steinmetz solid. If you intersect two cylinders with radius $r$ whose axes intersect perpendicularly, the volume is known to be $\frac{16}{3} r^3$. — achille hui, Jul 17 '16 at 23:01
This comment is to link this post as one of the (abstract) duplicates to the current choice of mother/target post. — Lee David Chung Lin, Jan 22 '19 at 10:29

Behrouz Maleki · Accepted Answer · 2016-07-17T23:04:38.877

2

The region of integration is even with respect all variables, thus $$V=8\int_{0}^{2}\int_{0}^{\sqrt{4-z^2}}\int_{0}^{\sqrt{4-z^2}}dxdydz$$ $$V=8\int_{0}^{2}\int_{0}^{\sqrt{4-z^2}}\sqrt{4-z^2}dydz$$ $$V=8\int_{0}^{2}(4-z^2)dz$$ $$V=8(4z-\frac{1}{3}z^3)|_{0}^{2}=\frac {128}{3} $$

edited Jul 17 '16 at 23:04

answered Jul 17 '16 at 22:59

Behrouz Maleki

11,342

Thanks Behrouz. Can you, or anyone, explain why I don't need to consider the projection in the xy plane here? – Ian Jul 18 '16 at 00:03
1

As Doug M said, you can consider the projection in the $xy$ plane, but if you consider the region ${(x,y);|;0\le x\le 2, 0\le y\le x}$, than $z$ ranges from $0$ to $\sqrt{4-x^2}$ (and not $\sqrt{4-y^2}$). – Kuifje Jul 18 '16 at 01:59
1

Please Note $-\sqrt{4-z^2}\le x \le \sqrt{4-z^2}$ and $-\sqrt{4-z^2}\le y\le +\sqrt{4-z^2}$. Since region of integration $D={(x,y,z)|-\sqrt{4-z^2}\le x \le \sqrt{4-z^2},-\sqrt{4-z^2}\le y\le +\sqrt{4-z^2}, -2\le z\le 2}$ is even with respect all variables, thus $\iiint_{D}dV=8\iiint $ – Behrouz Maleki Jul 18 '16 at 07:52

the volume of the solid limited by the surfaces $x^2+z^2=4$ and $y^2+z^2=4$.

1 Answers1