Solid of solids, pretty sure I got the wrong answer, not sure what I did wrong.

[saruji]

The file is a PDF, but here is an imgur link, anyone?

 

[Mark44]

The file is a PDF, but here is an imgur link, anyone?

That is incorrect, because your integral is not set up correctly. You are using horizontal disks (washers), but because of the shape of the region, their formula needs to change at y = 1/2. Between 0 and 1/2, the washers have the same outside diameter. Between 1/2 and 1, the washers have a different outside diameter. Since the formulas are different, you will need two integrals.

I would probably be inclined to use shells rather than washers in this problem.

 

[saruji]

So what would be my limits of integration than?...Having a hard time picturing this for some reason...

 

[iRaid]

Lol, I had this exact same problem on my quiz last week and got it wrong. I figured it out after class (while driving home, unfortunately) and what you have to do is set up 2 integrals. One from 0 to 1/2 and the other from 1/2 to 1. Adding these together gives you the total area.

$$\pi\int_0^.5 (outer)^{2}-(inner)^{2} dy + \pi\int_.5^1 (outer)^{2}-(inner)^{2} dy$$

That should make it much easier.

Edit: Just to add, you can tell yours is wrong when you go to evaluate it, what's ln 0? undefined..

 

[saruji]

Lol, I had this exact same problem on my quiz last week and got it wrong. I figured it out after class (while driving home, unfortunately) and what you have to do is set up 2 integrals. One from 0 to 1/2 and the other from 1/2 to 1. Adding these together gives you the total area.

$$\pi\int_0^.5 (outer)^{2}-(inner)^{2} dy + \pi\int_.5^1 (outer)^{2}-(inner)^{2} dy$$

That should make it much easier.

Edit: Just to add, you can tell yours is wrong when you go to evaluate it, what's ln 0? undefined..

Thank you so much