How to find the volume?

[helenli89]

Questions and solutions are attached.
I used a different method than the solution, but I should get the same answer; I don't know where it went wrong.

 

[helenli89]

I'm not so sure you seem to get a very different equation with the Spherical coordinates compare to the solution given by my professor who also uses the spherical coordinates. I don't think that with the same coordinates there could be 2 complete different equation to solve the same volumn. Otherwise we wouldn't have to memerize the volumn for cylinders when we were in high school.

I think My volumn for the sphere and the cylinder is incomplete and the 4pi/3 should be -4pi/3 (I made a misstake). I have found an example my prof did in class (attached). I only did the volumns for the part that's bounded by the shperes on top of the cylinder so I need to find the volumn for the part that's bounded by the cylinder but if I found the the hight of the cylinder then this become very easy. So I did the following. since x^2 +y^2 =1 -> x^2 +y^2 +z^2=2 becomes 1+z^2=2 then z^2=1 then half of the volum has hight 1 then the positive half of the cylinder have pi*r^2*1 = pi then the volumn that's bounded by the cylinder is 2pi the the whole thing is 2*pi-4pi/3. Then use the volumn of the sphere - this whole volumn which still doesn't equal to the correct answer.
Can anyone try this?

 

[BigGlenntheHeavy]

$\displaystyle What!! \ V \ = \ \frac{\pi h^{3}}{6} \ = \ \frac{\pi 2^{3}}{6} \ = \ \frac{4\pi}{3}.$

$\displaystyle Note: \ the \ above \ is \ done \ with \ x^{2}+y^{2}+z^{2} \ = \ 2, \ not \ 4, \ ergo, \ the \ radius \ is \ \sqrt2, \ not \ 2.$

 

[BigGlenntheHeavy]

$\displaystyle What, \ cat \ got \ your \ tongue, \ Helen?$

$\displaystyle Instead \ of \ going \ through \ all \ your \ machinations, \ Just \ used \ V \ = \ \frac{\pi h^{3}}{6}$

$\displaystyle A \ hole \ is \ cut \ through \ the \ center \ of \ a \ sphere \ of \ radius \ r.$

$\displaystyle The \ height \ of \ the \ remaining \ spherical \ ring \ is \ h.$

$\displaystyle Show \ that \ the \ volume \ of \ the \ ring \ is \ V \ = \ \frac{\pi h^{3}}{6}.$

$\displaystyle Note \ that \ the \ volume \ is \ independent \ of \ r.$

 

[helenli89]

$\displaystyle What, \ cat \ got \ your \ tongue, \ Helen?$

$\displaystyle Instead \ of \ going \ through \ all \ your \ machinations, \ Just \ used \ V \ = \ \frac{\pi h^{3}}{6}$

$\displaystyle A \ hole \ is \ cut \ through \ the \ center \ of \ a \ sphere \ of \ radius \ r.$

$\displaystyle The \ height \ of \ the \ remaining \ spherical \ ring \ is \ h.$

$\displaystyle Show \ that \ the \ volume \ of \ the \ ring \ is \ V \ = \ \frac{\pi h^{3}}{6}.$

$\displaystyle Note \ that \ the \ volume \ is \ independent \ of \ r.$

I can not use the formular! (read my posting carefully!) otherwise I won't have a problem with it at all!
anyways, I already know what's the problem.

 

[BigGlenntheHeavy]

$\displaystyle Helen, \ is \ it \ x^{2}+y^{2}+z^{2} \ = \ 2 \ or \ x^{2}+y^{2}+z^{2} \ = \ 4?$