Hi!! I need help with this problem :c

[Lester221]

Hello I'm doing the multivariable calculus homework and I'm stuck with this exercise:

Find the volume of $$\Omega= \left \{ (x,y,z)/ z\geq \frac{\sqrt{x^{2}+y^{2}}}{\sqrt{3}}, x^2+y^2+z^2\leq 2z, x^2+y^2\geq\frac{1}{4} \right \}$$
Making an intersection between surfaces I got the following values:
$$\left\{\begin{matrix} x^2+y^2+z^2= 2z\\ x^2+y^2=\frac{1}{4} \end{matrix}\right.\\ z=1+\frac{\sqrt{3}}{2},z=1-\frac{\sqrt{3}}{2},$$$$\left\{\begin{matrix} z= \frac{\sqrt{x^{2}+y^{2}}}{\sqrt{3}}\\ x^2+y^2+z^2= 2z \end{matrix}\right.\\ z=0,z=\frac{1}{2}$$
$$\left\{\begin{matrix} z= \frac{\sqrt{x^{2}+y^{2}}}{\sqrt{3}}\\ x^2+y^2=\frac{1}{4} \end{matrix}\right.\\ z=0,z=\frac{1}{2{\sqrt{3}}}$$

The answer muts be $$\frac{\pi}{4}(5-\sqrt{3}$$

But I got this $$\frac{1}{6}-\frac{\sqrt{3}}{8}$$


And and this is are my sketch on Geogebra


















I hope you guys can help me :c

 

[mmm4444bot]

But I got this $\frac{1}{6}–\frac{\sqrt{3}}{8}$

Please share your work, so that tutors may see what you've tried and check your work. Thank you!

Posting Guidelines (Summary)

Welcome to our tutoring boards! :) This page summarizes main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to help...

www.freemathhelp.com

$\;$

 

[blamocur]

And why did you bother to compute the values of $z$ ? Do you know what they correspond to geometrically ?

 

[Lester221]

Hello I'm doing the multivariable calculus homework and I'm stuck with this exercise:

Find the volume of $$\Omega= \left \{ (x,y,z)/ z\geq \frac{\sqrt{x^{2}+y^{2}}}{\sqrt{3}}, x^2+y^2+z^2\leq 2z, x^2+y^2\geq\frac{1}{4} \right \}$$
Making an intersection between surfaces I got the following values:
$$\left\{\begin{matrix} x^2+y^2+z^2= 2z\\ x^2+y^2=\frac{1}{4} \end{matrix}\right.\\ z=1+\frac{\sqrt{3}}{2},z=1-\frac{\sqrt{3}}{2},$$$$\left\{\begin{matrix} z= \frac{\sqrt{x^{2}+y^{2}}}{\sqrt{3}}\\ x^2+y^2+z^2= 2z \end{matrix}\right.\\ z=0,z=\frac{1}{2}$$
$$\left\{\begin{matrix} z= \frac{\sqrt{x^{2}+y^{2}}}{\sqrt{3}}\\ x^2+y^2=\frac{1}{4} \end{matrix}\right.\\ z=0,z=\frac{1}{2{\sqrt{3}}}$$

The answer muts be $$\frac{\pi}{4}(5-\sqrt{3}$$

But I got this $$\frac{1}{6}-\frac{\sqrt{3}}{8}$$


And and this is are View attachment 35510my sketch on GeogebraView attachment 35509


















I hope you guys can help me :c

I supposed i might extended this more (¿?)

Each Z values represent values of the intersection of the surfaces. Its mean, Spheres, Cylinder and The Cone I will add this sketch from Geogebra



Now, I'm working with this inequialities $$\Omega= \left \{ (x,y,z)/ z\geq \frac{\sqrt{x^{2}+y^{2}}}{\sqrt{3}}, x^2+y^2+z^2\leq 2z, x^2+y^2\geq\frac{1}{4} \right \}$$. t's mean a region outside the cylinder, inner from the sphere and inner from the cone.

So, for me the best option was this new region:



The red region is the cilynder region that i must ignore i suppose. And the withe region it's the region I must integrate, so I made use of the spherical coordinates and i got this:

$$\int_{0}^{2\pi}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\int_{\frac{sec\phi}{2\sqrt{3}}}^{\frac{sec\phi}{2}}\rho sec\phi d\rho d\phi d\theta$$


Thoses limits come from this:
$$\left\{\begin{matrix} x^2+y^2=\frac{1}{4}\\ z=\frac{1}{2}\\ z=\frac{1}{2\sqrt{3}}\\ x= \rho \cos\theta \sin\phi \\ y= \rho \sin\theta \cos\phi \\ z= \rho \cos\phi \\ z= \frac{\sqrt{x^{2}+y^{2}}}{\sqrt{3}} \end{matrix}\right.$$
But something still missing cuz i can't get the true answer :/

 

[blamocur]

Your answer cannot be correct if only because it is negative. But I have a strong suspicion that $\frac{\pi}{4}\left(5-\sqrt{3}\right)$ is not a correct answer either, although it is very close numerically: it differs from my Monte-Carlo simulation (with $2\times 10^8$ samples) by about 0.3%, whereas the answer I got differs by about 0.013%, i.e. my error is around 20 times smaller.

 

[Lester221]

Your answer cannot be correct if only because it is negative. But I have a strong suspicion that $\frac{\pi}{4}\left(5-\sqrt{3}\right)$ is not a correct answer either, although it is very close numerically: it differs from my Monte-Carlo simulation (with $2\times 10^8$ samples) by about 0.3%, whereas the answer I got differs by about 0.013%, i.e. my error is around 20 times smaller.

I expanded more in a reply I'm waiting for the moderator approval but yes, i got a negative volume that's the second problem too

 

[blamocur]

I expanded more in a reply I'm waiting for the moderator approval but yes, i got a negative volume that's the second problem too

Where does the "official" answer come from?

 

[blamocur]

And the withe region it's the region I must integrate, so I made use of the spherical coordinates...

Personally I find cylindrical coordinates more useful for solving this problem.

I also find your graphs difficult to understand. Since there is rotational symmetry around Z a 2D plot of the intersections of the surfaces with the XZ (or YZ) plane helped me to write the integral expressions.

 

[Lester221]

Where does the "official" answer come from?

It came from my teacher

 

[nasi112]

Hi there.

The idea of this problem is very simple. You have a cone and a sphere and you want to find the volume between them.

Find the volume of the full sphere, then subtract from it the volume between the cone and the bottom part of the sphere.

$\displaystyle x^2 + y^2 + z^2 \leq 2z$

This is the equation of the sphere, its center shifted 1 unit above the origin. If you solve for $\displaystyle z$, you will get

$\displaystyle 1 - \sqrt{1 - x^2 - y^2} \leq z \leq 1 + \sqrt{1 - x^2 - y^2}$

These are the equations of the top and bottom parts of the sphere.

As blamocur mentioned, solving this problem in Cylindrical Coordinate is straightforward.

The volume of the full sphere (removing some volume of the sphere as following the restrictions $\displaystyle x^2 + y^2 \geq 1/4$)

$\displaystyle \int_{0}^{2\pi}\int_{1/2}^{1}\int_{1-\sqrt{1 - r^2}}^{1+\sqrt{1 - r^2}} \ r \ dz \ dr \ d\theta$

The volume that you don't need is

$\displaystyle \int_{0}^{2\pi}\int_{1/2}^{1}\int_{1-\sqrt{1 - r^2}}^{r/\sqrt{3}} \ r \ dz \ dr \ d\theta$

Why is your teacher answer is a little bit different from my answer?

I think that your teacher by mistake solved the second integral as

$\displaystyle \int_{0}^{2\pi}\int_{1/4}^{1}\int_{1-\sqrt{1 - r^2}}^{r/\sqrt{3}} \ r \ dz \ dr \ d\theta$

$\displaystyle \int_{0}^{2\pi}\int_{1/2}^{1}\int_{1-\sqrt{1 - r^2}}^{1+\sqrt{1 - r^2}} \ r \ dz \ dr \ d\theta - \int_{0}^{2\pi}\int_{1/4}^{1}\int_{1-\sqrt{1 - r^2}}^{r/\sqrt{3}} \ r \ dz \ dr \ d\theta \ = 2.57 \approx \frac{\pi}{4}(5 - \sqrt{3})$