TITLE: Intuition of integration
QUESTION [2 upvotes]: While trying to derive the volume of a spherical cap, I noticed that my method of integrating small discs($\int \pi r^2 \, dr$) was wrong, since it resulted in the final formula $\frac{\pi r^3}{3}$, where r is the radius of the cap face.
Edit: this's what I did; $$\int \pi r^2 \, dr= \pi \frac{(r^3-0)}{3}=\frac{\pi R^3}{3}$$
If this was used to find the full volume of a sphere, it'd come up short, at $\frac{2\pi R^3}{3}$. This seems to be because integration seems to take a linear sum of values. For example, if you calculate the lower hemisphere volume using my incorrect method, the result will essentially only be that of a cone with a vertice at the bottom, since the hemisphere radius increases at a faster rate than the linear rate of the cone radius.
What is the intuitive reason for integration being a linear sum in this way?

REPLY [1 votes]: In your method, you are finding the area of a parabola $y=\pi r^2$ from $r=0$ to $R$. [I am assuming that you meant the radius $R$ of the sphere instead of $r$ in your answer.] Note, that if this were valid, it would be the volume of the sphere, not the volume of the cap. You are using $r$ as a variable whereas, evidently it is supposed to be the fixed radius of the cap face. To find the volume of the cap face, you should be 'summing' the varying areas of the cross-sections parallel to the face of the cap.
If $r$ is the fixed radius the face of a cap of a sphere of radius $R$ and you want to compute the volume of the cap by integrating the cross-sections parallel to the fixed face, then you need to integrate with respect to some variable $x$ measured along the central axis of the cap. For each $x$ you will have some $y$ corresponding to the radius of the cross section of the cap parallel to the face.
Then your integral will be expressed as
$$ \int \pi y^2\,dx $$
You have enough information to find the limits of integration in terms of $R$ and $r$ and $y$ in terms of $x$ and $R$.
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