I want to calculate the volume of a sphere with radius $r$ using calculus. I can split this sphere into lots of great circles of that sphere and then add all the areas to find the volume. Let's write

$$ V = \int_{0}^{\pi} \pi r^2 d \theta = \pi^2r^2 $$

But we all know the volume should be $\frac{4}{3} \pi r^3$. So, what's wrong with this?


You actually want to calculate $\int_{-r}^r \pi(r^2-x^2) dx$, where by the Pythagorean theorem $r^2-a^2$ is the squared radius of the circle at $x=a$. In polar coordinates, another option is $\int_0^{2\pi}d\phi\int_0^\pi \sin\theta d\theta\int_0^r \rho^2 d\rho$.

  • $\begingroup$ I know this is correct. But I "think" my solution should lead to another correct solution. I know other ways to solve this but my question is why my solution is not correct. $\endgroup$ – Ehsan Poursaeed Jun 2 '18 at 14:18
  • 1
    $\begingroup$ @EhsanPoursaeed See my revised answer for something closer to your approach. You didn't get the Jacobian right. $\endgroup$ – J.G. Jun 2 '18 at 14:19
  • $\begingroup$ I don't know what your second solution is :(. Can you explain it? $\endgroup$ – Ehsan Poursaeed Jun 2 '18 at 14:20
  • $\begingroup$ @EhsanPoursaeed You might want to start by understanding why the circle's area is $\int_0^{2\pi}d\theta\int_0^r \rho d\rho$. That $\rho$ is a Jacobian. In the 3D case, it's $\rho^2\sin\theta$. $\endgroup$ – J.G. Jun 2 '18 at 14:30

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