Using solids of revolution to find the volume of a sphere cap

Question

I want to find the volume of a sphere cap using the solids of revolution method. Let the sphere have radius $r$ and the cap have height $h$. Then the volume of the cap is given by $\pi h^2\left(r-\frac{1}{3}h\right)$.

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I tried to derive the expression as follows: the equation for a circle with centre $(0,0)$ is $x^2+y^2=r^2\implies x=\sqrt{r^2-y^2}$ So the volume = $$\int\limits_{r-h}^r \pi(r^2-y^2)\operatorname{d}y=\left[\pi r^2y-\frac{1}{3}\pi r^2y^3\right]_{r-h}^r=\pi r^2h-\frac{1}{3}\pi r^2h^3$$ However, this solution is wrong. Where is the error in my working?

I have asked this question before here and here, in similar posts, but have attracted no feedback.

Answer 1

The source of error is in your integration.

The integration should yield the following result $$\int\limits_{r-h}^r \pi(r^2-y^2)\operatorname{d}y=\left[\pi r^2y-\frac{1}{3}\pi y^3\right]_{r-h}^r=\pi rh^2-\frac{1}{3}\pi h^3$$

Once you've corrected this part you'll get the right answer.

Answer 2

$$\int\limits_{r-h}^r \pi(r^2-y^2)\operatorname{d}y=\left[\pi r^2y-\frac{1}{3}\pi r^2y^3\right]_{r-h}^r$$

Why did you carry $r^2$ into the integral once again?

When carrying out computations of length, area and volume, check that the degree of the integrand is in fact 1,2 or 3 respectively. This is dimensional check from physics and you often benefit in your work by dimensional check.

The moment you noticed that your right hand side had a fifth degree dimension term instead of third for volume, look back then and there without going forward with expensive computations and cross out $r^2$ term sooner than later. Sorry for the emphasis.. consequent to your posting repeatedly regarding such detail.