I was given the following question:
Given a cylinder circumscribed within a parallelepiped with a square base that has a plane going through the center of the base circle and through one side of the square on the top of the parallelepiped. Use calculus to prove that the volume of the segment of the is $\frac16$ the volume of the rectangular parallelepiped circumscribing the cylinder.
Here is what I've done:
I created a diagram of the described cylinder on GeoGebra (see below). Given that the base is a square, the volume of the parallelepiped can be given as $a^2b$ (if necessary this can be shown using calculus by doing $\int^a_0 ab\ dx$). The radius of the cylinder is $\frac a2$ and the height is $b$. The volume would therefore be: $\frac{\pi a^2}{4}b$. The only problem is that I need to factor in the plane. I would take the integral of the equation of the plane minus the area of each cross section of the cylinder ($\frac{\pi a^2}{4}$) but I cannot figure out the equation of the plane! I know it goes through the point $(\frac a2,\frac a2,0)$ and the line $z=b,x=a$ (I'm not sure if I wrote that line right). But I don't know how to get the equation of that plane.
Am I going about this problem in the right way? How can I find that equation?
Let the cylinder be $x^2+ y^2=\frac{a^2}4$. Then, the corresponding cutting plane is $z=\frac {2b}a x$ and the section volume is
$$\int_ {-\frac a2} ^{\frac a2} \int_0^{\sqrt{\frac{a^2}4-y^2} }z \>dxdy = \int_{-\frac a2} ^{\frac a2} \int_0^{\sqrt{\frac{a^2}4-y^2} }\frac {2b}a x\> dxdy = \frac16 a^2b$$ which is $\frac16$ that of the rectangular parallelepipe.