# Thread: Sphere volume derivation with the correct variable of integration

1. ## Sphere volume derivation with the correct variable of integration

Hello all,

So I'm trying to figure out why setting up integrals to find volume of a sphere with the correct variable of integration.

For example, for volume,

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \pi r^2 \cos^2(\theta)d\theta=\frac{\pi r^2}{4}$ does not work.

I believe my teacher explained that it had to do with the $d\theta$ that were in fact small wedges that, in sum, did not produce a sphere. However, the substitution using $x=r*\sin(\theta)$ and $dx=r*\cos(\theta)d\theta$ did produce the correct result, which is actually the same as the conventionally derived (single-variable) shells integral for a sphere:

$\pi \int_{-r}^{r} x \sqrt{r^2-x^2} dx$

Now, shouldn't these integrals be the same, despite the substitution? What was the first integral really solving for- I don't think I can picture it in my head.

Thank you for taking the time to read this post and help me out.

2. Top integral should be cos^3 not squared and you forget a r. Those two integrals aren't the same otherwise.

And the volume of a sphere is $\displaystyle \frac{4\pi r^3}{3}=\pi r^3\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\cos^3{(\theta)}d\theta$

3. Actually, you may be right. Do you by any chance know the proper substitution? The first integral was essentially a "washers" integral using $d\theta$ as the variable of integration. I'm actually not sure if is the washers or shells $dx$ integral that results after the substitution.

I changed the 2nd integral in my post to the shells integral, btw.

Thanks so much for helping out!

4. Shell method is of the form:

$\displaystyle 2\pi\int_{a}^{b}p(x)h(x)dx$

5. So I forgot another $r*\cos^(\theta)$ term! But how? I thought I had the general form of shells correct, just:
$\int_{\theta_1}^{\theta_2} \pi r(\theta) d\theta$

and in the 1st integral's case, $r(\theta)=r*\cos(\theta)$, no?

Thanks again.

EDIT: I meant washers, sorry.

6. EDIT to post below: I meant washers for the 1st integral, not shells, sorry.

7. Everything was correct except for forgetting the $r*\cos(\theta)$.

If you would do the integration with $r*\cos(\theta)$, you would obtain your desired results.

8. Shell with x=rsin substitution

$\displaystyle 4\pi r^3\int_{0}^{\frac{\pi}{2}}(\sin{\theta}*\cos^2{\t heta})d\theta=\frac{4\pi r^3}{2}$

Washer with x=rsin substitution

$\displaystyle \pi r^3\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\cos^3{\theta}d\theta=\fra c{4\pi r^3}{3}$

9. Hmmm... I distinctly remember my 1st integral the one we used in class and it worked, but I think you are showing me what I should do for the shells method. I think I need the washers method actually.

EDIT: I just saw your reply above, thanks for helping me understand shells using $d\theta$ so much better. But I was wondering if you could still show me washers- thats the one I initially wanted.

10. Originally Posted by progressive
Hmmm... I distinctly remember my 1st integral the one we used in class and it worked, but I think you are showing me what I should do for the shells method. I think I need the washers method actually.

EDIT: I just saw your reply above, thanks for helping me understand shells using $d\theta$ so much better. But I was wondering if you could still show me washers- thats the one I initially wanted.
I edited my post #8 with both methods and with your desired substitutions.

11. I think I get it now. Thanks very much for all your help!