# Thread: Volume of sphere minus cylinder

1. ## Volume of sphere minus cylinder

The sphere has a radius of 5

The cylinder has a radius of 4

My integral for my circle is $\int_0^5(25-x^2)dx$

My integral to subtract from that is $\int_0^4(25-x^2)dx$

I do not think this is right

any help?

2. Originally Posted by silencecloak
The sphere has a radius of 5

The cylinder has a radius of 4

My integral for my circle is $\int_0^5(25-x^2)dx$

My integral to subtract from that is $\int_0^4(25-x^2)dx$

I do not think this is right

any help?
First think of this in the upper xy-plane.
The equation of the top sphere is $z=\sqrt{25-x^2-y^2}$.
If you compute, $\iint_R \sqrt{25-x^2-y^2} ~ dA$ where $R = \{ (x,y) | x^2+y^2 \leq 16 \}$ you get the area of the cylinder below sphere.
Thus, the total volume this cycliner-like piece takes up is double that integral.
Once you know this number you can finish the problem.

3. Originally Posted by ThePerfectHacker
First think of this in the upper xy-plane.
The equation of the top sphere is $z=\sqrt{25-x^2-y^2}$.
If you compute, $\iint_R \sqrt{25-x^2-y^2} ~ dA$ where $R = \{ (x,y) | x^2+y^2 \leq 16 \}$ you get the area of the cylinder below sphere.
Thus, the total volume this cycliner-like piece takes up is double that integral.
Once you know this number you can finish the problem.
I can kind of see how that works but I would never come up with something like that myself(And honestly I don't know where to start to compute that). The way we were taught in class is how I have tried to set it up.

$2\pi\int_0^5(25-x^2)dx - 2\pi\int_0^4(25-x^2)dx = 29.3215314335$

or

$2\pi\int_4^5(25-x^2)dx = 29.3215314335$
Would this be correct?

4. Consider $y=\sqrt{25-x^2}$ the top half of the circle.
If you compute, $2\pi \int_{-3}^3 xy dx$ you get the volume of the cyclinderical piece.

5. I got it today in class, thank you for your help...

Problem was i was trying to rotate the shape around the wrong axis

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### volume of a sphere minus cylindric integral

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