Question regarding the volume integral of a 3D solid.

Question

$\mathbf {The \ Problem}$: Find the volume of the solid in $3D$ enclosed by the three cylinders with their equations given below,

$$x^2+y^2 =1,$$

$$ y^2+z^2 =1, $$

$$ z^2+x^2 =1 .$$

$\mathbf {My \ approach}$: Actually, I tried to figure it out using the polar coordinates. Here, $0\leq r\leq 1$ and

$$z = \min (1-r^2\operatorname{sin^2\theta}, 1-r^2\operatorname{cos^2\theta} ).$$

Hence the volume is

$$V = 4 \left( \int_0^{π/4} \int_0^1 \sqrt{1-r^2\cos^2\theta} rdrd{\theta} + \int_{π/4}^{π/2} \int_0^1 \sqrt{1-r^2\sin^2\theta} rdrd{\theta} \right)$$

then the answer is found to be $8/(3\sqrt{2})$ .

there was an exact same question a few hours ago, and apparently this link helped that user https://math.stackexchange.com/questions/923508/calculating-volume-enclosed-using-triple-integral Also, I guess part of the problem is that you're missing a whole bunch on square-roots. — peek-a-boo, Aug 24 '19 at 04:56
This region has more symmetry of which you can take advantage and get a very simple integral. In fact, the problem can be solved, using the way Archimedes would have, without using calculus at all. — Ted Shifrin, Aug 24 '19 at 19:20
@TedShifrin, Sir, can you please explain me about Archimedes's solution or provide me a link ??? — Rabi Kumar Chakraborty, Aug 24 '19 at 19:30
Take a look at pages 2-9 of a lecture I gave some years ago. — Ted Shifrin, Aug 24 '19 at 20:02

Quanto · Accepted Answer · 2019-08-25T03:48:24.197

0

Your volume integral is fine, except for an overall factor of 8 instead of 4.

The integral below is expressed in $xy$-coordinates in the first quadrant, for $z > 0$. The factor 8 accounts for 8 identical integral contributions to the total volume.

$$V = 8 \int_0^{π/4} \int_0^1 \sqrt{1-r^2\cos^2\theta} rdrd{\theta} + 8\int_{π/4}^{π/2} \int_0^1 \sqrt{1-r^2\sin^2\theta} rdrd{\theta} $$

The two integrals in the above expression are actually the same in value due to symmetry between sine and cosine.

Thus,

$$V= 16 \int_0^{π/4} \int_0^1 \sqrt{1-r^2\cos^2\theta} rdrd{\theta} = 8(2-\sqrt{2})$$

edited Aug 25 '19 at 03:48

answered Aug 24 '19 at 15:48

Quanto

97,352

Sir, how can you tell that the result in the link is incorrect ??? – Rabi Kumar Chakraborty Aug 24 '19 at 19:31
Double checked the last step, which is actually equivalent to the last couple of lines of the link. So, your integral gives the same result. Sorry for the confusion. – Quanto Aug 25 '19 at 03:45

poetasis · Answer 2 · 2019-08-24T19:36:33.420

0

You have three equations of circles which is the same as one equation for a sphere. Here is a link showing the integration steps needed to find it. In your case, with a radius of $1$, the formula $\frac{4\pi r^3}{3}$ reduces to $$V=\frac{4\pi}{3}$$

edited Aug 24 '19 at 19:36

answered Aug 24 '19 at 19:12

poetasis

6,338

Question regarding the volume integral of a 3D solid.

2 Answers2