# Can't figure my mistake in calculating volume

Can't figure out my mistake in calculating volume

I'm trying to calculate the volume of {(x,y,z)|$x^2+(y-1)^2<1$, $0<z<\sqrt{4-x^2-y^2}$}:

$$\int_0^{\pi}d\theta \int_0^{2sin(\theta)}\sqrt{4-r^2}rdr=\frac {-1}{3}\int_0^\pi ((4-4sin^2\theta)^{3/2}-8)d\theta=\frac{-8}{3}(\int_0^\pi (1-sin^2\theta)\sqrt{1-sin^2\theta}d\theta-\pi)=\frac{-8}{3}(\int_{sin(0)}^{sin(\pi)} (1-u^2)du-\pi)=\frac{8\pi}{3}$$

However on Wolfram Alpha I get different answer for the same integral.
What I'm doing wrong?

Last edited:

I like Serena
Homework Helper
Hi estro!

You seem to have integrated ##\sqrt{4-r^2}## as ##{-1 \over 3}(4-r^2)^{3/2}##.
But that is not its primitive.
You can see this if you calculate the derivative, applying the chain rule.

Btw, I have the impression that your integral does not represent the volume that it's supposed to.
Your volume is a cylinder bounded asymmetrically by a sphere.
How does that map to your integral?

Hi and thanks for the quick answer.
Indeed I made a mistake in writing the integral, it should be $$\int_0^{\pi}d\theta \int_0^{2sin(\theta)}\sqrt{4-r^2}rdr$$

Fixed in original post as well.

I like Serena
Homework Helper
Okay... I still don't see how that maps to your volume...

And it looks like you substituted u=sinθ.
Did you consider that cosθ does not have the same range as the square root?

I like Serena
Homework Helper
Up to you.

But your mistake appears to be that cosθ becomes negative for θ>pi/2.
You will need to split your final integral in a part where cosθ is positive and a part where it is negative.

Okay... I still don't see how that maps to your volume...

And it looks like you substituted u=sinθ.
Did you consider that cosθ does not have the same range as the square root?

I'm looking for the volume which is under top half ball of radius 2 and above circle x^2+(y-1)^2=1

indeed I used u=sinθ so the limits of integration became [itex]\int_{sin(0)}^{sin(\pi)} so the integral becomes 0 which is strange, however I can't pinpoint where is the problem.

Up to you.

But your mistake appears to be that cosθ becomes negative for θ>pi/2.
You will need to split your final integral in a part where cosθ is positive and a part where it is negative.

Oh, maybe it is it. Let me check...

I like Serena
Homework Helper
As I said the square root equals |cosθ|, while you assumed it equals cosθ.

Btw, substitutions of any kind are always the trickiest parts in calculations.
Hint: always do substitutions without doing anything else (no matter how cumbersome that seems to be at first).
That makes it easier to do them right and it makes it especially easier to track mistakes (also for us ;)).
After a substitution you can make a step as big as you want.

Ok so here is the problem:

$$\int_0^\pi sin^2\theta \sqrt{1-sin^2\theta}d\theta=\int_0^\pi sin^2\theta cos\theta d\theta=\int_{sin(0)}^{sin(\pi)} u^2du=0$$

Seems like this is wrong, but why?

I like Serena
Homework Helper
Try:
$$\int_0^\pi \sin^2 θ \sqrt{1-\sin^2 θ}\ dθ = \int_0^\pi \sin^2 θ |\cos θ|\ dθ = \int_0^{\pi/2} \sin^2 θ \cos θ\ dθ + \int_{\pi/2}^{\pi} \sin^2 θ \cdot -\cos θ\ dθ$$

Try:
$$\int_0^\pi \sin^2 θ \sqrt{1-\sin^2 θ}\ dθ = \int_0^\pi \sin^2 θ |\cos θ|\ dθ = \int_0^{\pi/2} \sin^2 θ \cos θ\ dθ + \int_{\pi/2}^{\pi} \sin^2 θ \cdot -\cos θ\ dθ$$

You're right, I get it now, HUGE thanks!
I'm pretty much used to do the substitutions in auto mode without thinking, sometimes it comes costly...

Thanks again!