Thread: Getting a negative volume using shell method

Hello!
I hope you don't mind that I took a picture of my work since there is a lot written down.

The meat of the matter is that I am getting a negative volume as my answer when I use the shell method, but not with the disk method.

In my opinion, the problem arises during integration using trig substitution and then distributing the negative from the original integration by parts formula. I have tried building the triangle making both sin and cos = y.

Maybe it's just something simple I'm not seeing, like order of operations error, but it's driving me crazy! Thanks in advance.

The mistake I see is you integrated $\displaystyle \int cos(2\theta) d\theta$ wrong, it should be $\displaystyle \frac{1}{2}sin(2\theta)$ because $\displaystyle \frac{d}{d\theta}[\frac{1}{2}sin(2\theta)]=cos(2\theta)$ ignoring the constant of integration ...it looks like you have $\displaystyle \int cos(2\theta) d\theta=2cos(\theta)sin(\theta)$ as the result, which looks like the double angle formula for sine on the left side.

Is it wrong to substitute the double angle form of sin(2(theta)) in? Aren't they interchangeable?

Originally Posted by Vee

Is it wrong to substitute the double angle form of sin(2(theta)) in? Aren't they interchangeable?

oh, i see what you did now, you confused me, lol, you used the double angle after you integrated which is correct, back to the drawing board.

In the triangle you drew to define theta you have $\displaystyle cos(\theta)=\sqrt{1-y^2}$ so you can't have $\displaystyle \theta =cos^{-1} (y)$ right? Your expression around the bottom left should look like $\displaystyle \frac{-1}{4}sin^{-1}(y)-\frac{1}{4}y\sqrt{1-y^2}$ ...

EDIT: one more thing, it looks like you defined your $\displaystyle \theta$ more than once using two inconsistent triangles. I just noticed the second triangle, this is most likely the source of your error.

I thought about that. But the first triangle was just to find the derivative of sin^1(y). I also tried rebuilding my triangle for the trig substitution so it's the same as the triangle I used to differentiate inverse sin. However, it didn't matter, I still came out with the same value in the end. I think because inverse sin and inverse cosine only differ by a constant so integrating them makes them equal.

Here's how it works for me:

$\displaystyle 2\pi\left[\frac{1}{2}y^2sin^{-1}y+\frac{1}{4}sin^{-1}y+\frac{1}{4}y\sqrt{1-y^2}-\frac{\pi}{4}y^2\right]^1_0$

$\displaystyle =2\pi[(\frac{\pi}{4}+\frac{\pi}{8}+0-\frac{\pi}{4})-(0+0+0-0)]$

$\displaystyle =\frac{\pi^2}{4}$

How do you get positive (1/4)sin^-1(y)? When you distribute the minus sign from doing integration by parts doesn't it make it negative?

I found the problem. Inverse sin has the range x E [-pi/2, pi/2]. And our range is from pi/2 to pi. So by symmetry, we have to look at the volume from 0 to pi/2. That makes the height reversed pi/2 -sin^-1(y). That will change that negative into the wanted positive! Yay! Thank you so much for working this out with me. It has really helped me in thinking this though :-)

Originally Posted by Vee

How do you get positive (1/4)sin^-1(y)? When you distribute the minus sign from doing integration by parts doesn't it make it negative?

I used the principles of magic and changed the rules of algebra temporarily for our purposes. We needed a positive value in front of there so I figured the ends justify the means.

Originally Posted by Vee

I found the problem. Inverse sin has the range x E [-pi/2, pi/2]. And our range is from pi/2 to pi. So by symmetry, we have to look at the volume from 0 to pi/2. That makes the height reversed pi/2 -sin^-1(y). That will change that negative into the wanted positive! Yay! Thank you so much for working this out with me. It has really helped me in thinking this though :-)

Good thinking, I agree with this, my next line of attack was the original integral. I think your triangles and def. of theta confused me and I had convinced myself the error was the integration itself but now I'm supremely confident your integration was correct all along because I just worked through it in I get the same thing you did.

And you're welcome, I like helping when I can, sometimes a poster seeking help ends up helping me with his own problem.