I am learning Calculus from MIT lectures available on OCW. I am learning Calculus for the first time. So please be considerate if the question sounds stupid.

In the lecture on computing volumes using the disk method, the professor gave an example of computing the volume of a sphere of radius $a$ by rotating a circle centered at $(a,0)$ around the x-axis.

The professor computed the radius $y$ of the disk of revolution using the following equation for the circle

$(x-a)^2 + y^2 = a^2$

$\implies x^2 -2ax + y^2 = 0 $

$ \implies y^2 = 2ax - x^2 $

He then used this value of $y^2$ in the integration below

$\int_{0}^{2a} \pi (2ax - x^2) \, dx$

and ended up calculating the volume of the sphere to be $\frac{4}{3} \pi a^3$ which is correct.

My question is that if suppose say I wanted to calculate the volume of a solid of revolution obtained by just rotating the top half of the circle.

If I use the equation $y = +\sqrt {2ax - x^2} $ and then use the formula $\int_{0}^{2a} \pi y^2 \, dx $, I will end up calculating the same volume as calculated above for the sphere. What is the mistake here? Is the equation for the top semicircle that I got above not correct? Am I doing something wrong here?

What should I use as the value of $y$ if I need to compute the volume of the solid obtained by rotating the top half of the circle with radius $a$ centered at $(a,0)$

  • $\begingroup$ I am sorry. I just realized that the question was stupid to begin with. Didnt realize that the semicircle would end up creating the same sphere. Imagination Level SCORE - 0. Please ignore this question $\endgroup$ Commented Sep 24, 2020 at 8:29
  • $\begingroup$ If you don't want people to see this question, you can delete it $\endgroup$
    – DatBoi
    Commented Sep 24, 2020 at 9:03

1 Answer 1


Your top semicircle is OK but should be integrated from $a$ to $2a$ instead of what you did [from $0$ to $2a$ which gives the whole sphere].

  • $\begingroup$ Yeah just realized it. Thank you $\endgroup$ Commented Sep 24, 2020 at 8:31
  • $\begingroup$ You're welcome. $\endgroup$
    – coffeemath
    Commented Sep 24, 2020 at 8:32

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