I would like to compute the volume of the following cylindrical wedge 
(the portion in yellow) using an integral.
The plane that cuts the wedge goes through the very bottom of the cylinder leading to an ellipse as the cross section of the wedge.
The ellipse has major axis $R$ and minor axis $r$.
The ellipse's major axis makes an angle $\theta_{0}$ with the vertical axis/end of the cylinder. We have the following relations
$$R\cos\theta = r$$
$$h=R\sin\theta_{0}=r\tan\theta_{0}$$
where $h$ is the height of the cylindrical wedge.
How do I think through how to set up an integral for this volume?
Best Answer
If you cut with a plane from the tip of one end disk of the cylinder to the other end disk, then your problem is symmetrical (by rotating by a half-turn along the ellipse's minor axis). So basically compute the cylinder volume delimited by vertical sections at the endpoints of the ellipse and divide by $2$.
The volume of the cylinder is thus $$h \cdot \frac{\pi \cdot r^2}{4} = \big( R \cdot \sin(\theta_0) \big) \cdot \left(\frac{\pi}{4} \cdot R^2 \cdot \cos^2(\theta_0)\right) = \frac{\pi}{4} R^3 \sin(\theta_0)\cos^2(\theta_0)$$ And your final result is thus : $$ \frac{\pi}{8} R^3 \sin(\theta_0)\cos^2(\theta_0) $$
This is a "hack" taking advantage of symmetry. The consistent way is to integrate along $h$ the function of the area of a vertical section of the yellow volume, but it's way more tedious.