The sphere has a radius of 5
The cylinder has a radius of 4
My integral for my circle is $\displaystyle \int_0^5(25-x^2)dx$
My integral to subtract from that is $\displaystyle \int_0^4(25-x^2)dx$
I do not think this is right
any help?
The sphere has a radius of 5
The cylinder has a radius of 4
My integral for my circle is $\displaystyle \int_0^5(25-x^2)dx$
My integral to subtract from that is $\displaystyle \int_0^4(25-x^2)dx$
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 $\displaystyle z=\sqrt{25-x^2-y^2}$.
If you compute, $\displaystyle \iint_R \sqrt{25-x^2-y^2} ~ dA$ where $\displaystyle R = \{ (x,y) | x^2+y^2 \leq 16 \}$ 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.
$\displaystyle 2\pi\int_0^5(25-x^2)dx - 2\pi\int_0^4(25-x^2)dx = 29.3215314335$
or
$\displaystyle 2\pi\int_4^5(25-x^2)dx = 29.3215314335$
Would this be correct?
Thank you for your help