# Thread: triple integral - volume

1. ## triple integral - volume

hello,

i am very stuck with this pls help...

Write a triple integral in cylindrical coordinates for hte volume of the solid cut from a sphere of radius 2 by a cylinder of radius 1, one of whose rulings is a diameter of the sphere. Take the axis of teh cylinder parallel to the z axis.

okay so i have the equation for the sphere to be x^2 + y^2 + z^2 =2 replacing x=rcos(theta) and y=rsin(theta) I get z=+/- sqrt.root.of [2-r^2], which I fancy are the limits for the z variable

its projection onto the r,theta plane gives r<=cos(2theta) so I fancy that the limits for the r variable are 0 to cos(2theta)

and the limits for the theta variable from -pi/2 to pi/2

the problem is i am wrong, because the back of the book gives the mass of the solid to be "9,56 times the uniform density ro"

but I don't get as the volume 9,56

2. The equation for a sphere of radius $\displaystyle r$ is

$\displaystyle x^2+y^2+z^2=r^2.$

Our equation for a sphere will therefore be

$\displaystyle x^2+y^2+z^2=2^2=4.$

Hope this helps!

3. I don't get that either. The equation of the sphere is $\displaystyle x^2+y^2+z^2=4$. And if I take out a cylinder with radius of one, then the volume of that piece removed I calculate to be:

$\displaystyle V=2\sqrt{3} \pi +2\int_0^{2\pi}\int_0^1 \int_{\sqrt{3}}^{\sqrt{4-r^2}} r dz dr d\theta\approx 11.74$

The $\displaystyle 2\sqrt{3}\pi$ is just the cylinder part without the top and bottom part of the sphere. Then I calculate the volume of the top part of the sphere over the cylinder with the triple integral, then multiply by two for the bottom as well.

4. thanks so much

could you possibly me a bit more analytic about how you got the limits?

appreciate that

5. Try and figure it out based on the plot below. You can calculate just the volume of just the cylinder without the tops, then all we need is the volume of the green part right? You can calculate the length of the blue and red lines. Now take that circular area which is underneath the green part and place it onto the x-y plane. That is a circle right with diameter one. So you know how to integrate over a circle of radius one right? That's the 0 to 1 and 0 to 2pi bit. Now for the z limits, isn't that just from the bottom of the green part to the top of the green part as a function of the outward radius?

6. ## great!

Originally Posted by shawsend
Try and figure it out based on the plot below. You can calculate just the volume of just the cylinder without the tops, then all we need is the volume of the green part right? You can calculate the length of the blue and red lines. Now take that circular area which is underneath the green part and place it onto the x-y plane. That is a circle right with diameter one. So you know how to integrate over a circle of radius one right? That's the 0 to 1 and 0 to 2pi bit. Now for the z limits, isn't that just from the bottom of the green part to the top of the green part as a function of the outward radius?
I can't believe it! all this time i had stopped trying to solve the exercise and got back to the theory finished studying the chapter started again form exercise one and reached this exercise (12) again and i have the same number as you! i am absolutely chuffed, thanks for posting!

7. Originally Posted by shawsend
Try and figure it out based on the plot below. You can calculate just the volume of just the cylinder without the tops, then all we need is the volume of the green part right? You can calculate the length of the blue and red lines. Now take that circular area which is underneath the green part and place it onto the x-y plane. That is a circle right with diameter one. So you know how to integrate over a circle of radius one right? That's the 0 to 1 and 0 to 2pi bit. Now for the z limits, isn't that just from the bottom of the green part to the top of the green part as a function of the outward radius?
the limits i used are for z from -to + sqrt.root.of 4-r^2
for r form zero to one and for theta form 0 to 2pi