The region between the curve Y equal negative cosecant inverse of X and the line Y equal π halves from X equal 1 to X equal 2 divided by root 3 is revolved about the Y axis. To generate a solid, find the volume of the solid first thing. I don't like their picture, so I'm going to draw my own picture and I'm going to realize that this is my curve right here for Y equal negative cosecant inverse of X. So that's Y equal negative cosecant inverse X. And here's my Y equal π halves from X equal 1 to X equal 2 over root 3. So the first thing I want to do is find these points. So if I put in X equal 1, I get negative cosecant inverse of one which is just negative π halves. And if I put in two over root 3, the negative cosecant inverse of two over root 3 is negative π thirds. I'm also going to solve for X here. So I'm going to divide by a negative and get negative Y equal cosecant inverse of X. Take the cosecant of each side, and then knowing my even and odd identities, cosecant and negative Y is really just negative cosecant Y. So that's X equal negative cosecant Y. Now I'm going around this Y axis, so I'm going to do the washer method, which is π R-squared. And I'm going to do my outer R minus my inner R. So if we think about putting in line segments here, my outer R is going to be that equation. My inner R is going to be one because we're talking about a distance. So Pi, the bottom Y value negative π halves to negative π thirds of this equation here in terms of YS, because I'm going to do it dy if I think about stacking them this way. So we're going to have negative cosecant y ^2 -, 1 ^2 dy outer radius squared minus inner radius squared. And that's this little piece down here. So we're going to add to it π from negative π thirds up to π halves of the outer radius. In this case, the outer radius is going to be two over root 3, and we're going to square it minus the inner radius of one. Remember we're going DY so we're stacking these washers. So two over root 3 ^2 -, 1 ^2 d Y. So simplifying this up, the negative cosecant y ^2 is going to be cosecant squared y -, 1 two over root 3 ^2 is going to be 4 thirds -1 or 4 thirds -3 thirds or 1/3. So then we're going to have the negative cosequent y ^2, which is just going to give me positive cosequent squared Y. What gives? What's the integral of cosecant squared YDY? And that's going to be negative cotangent Y. The integral of -1 is going to be negative Y, negative π halves to negative π thirds 1/3 is going to turn into Pi 1/3 Y from our bounds. So then when we substitute in, we get π negative cotangent negative π thirds minus negative π thirds minus negative cotangent negative π halves minus negative π halves plus π thirds, Pi halves minus negative π thirds. And then it's just doing a bunch of manipulation algebra wise and we get π root 3 / 3 + π ^2 / 9.