Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Shell method. The volume equal A to B of 2π times the radius of the shell times the height of the shell times the thickness, traditionally DX or DY. So if we're going about a vertical line, IE something like the Y axis, we would have the volume equaling A to B of 2π shell radius times shell height times DX. In the shell method, we're always going to be parallel to the line we're going around as far as our DX or our DY. If we're going about a horizontal line like the X axis, we would be going A to B, or we could think of it as C to D if you prefer, of 2π. The shell radius times the shell height times DY. Now when we think about this, our shell radius is really going to be the Y here each radius around up here. Our shell radius is going to be the X each time we go out in a circle. So if we thought about Y equal to minus X ^2 / 4 going around the Y axis, we'd have 2π. Our radius of X0 to two was given of the height the highest 2 -, X ^2 / 4 minus the lowest. In this case, it was bounded by the X axis, so the X axis 0 less than or equal to X less than or equal to two. So then we're going to distribute and we're going to get two X -, X ^3 / 4 DX. Evaluate the integral and we get 6 Pi cubic units, so 2π. The radius was how far away we were from what we're going around the height, the highest Y value minus the lowest Y value. If we look at this one, we're given Y equal sqrt 3, X equal 3 -, y ^2 bounded in the first quadrant. So we're going to set the two and X equal 3. So we're going to set X equal 3 equaling 3 -, y ^2 so that we can come up with our bounds. Did they give, oh, they gave us our bounds 0, sorry, zero to root 3, root 3 up here X axis because we're in the 1st quadrant. SO0 to root 32 Pi. Our radius this time, if we're going around the X axis is the height up that we go each time. So the radius right here is our Y. Now we need to think about the line segment length. So it's going to be the furthest to the right minus the furthest to the left in terms of our X's. So 3 -, 3 -, y ^2 and that's all going to be DY. So if we distribute, we'd get three y -. 3 Y minus a negative is plus y ^3 d Y so 2π zero to root 3 Y cubed DY2 Pi 1/4 Y to the fourth. Putting in our upper bound minus our lower bound, root 3 to the 4th is just going to give us 9, so we'd get 9 Pi halves cubic units. If we look at the next one, Y equal 2 -, X ^2 y equal X ^2 X = 0 about the Y axis. So this time they didn't give us our bounds. We're going to set the two equations equal and solve so we can see that X is 1 and -1 If we're going around the Y axis, we want this to be DX, so we're going to have 2π. Our radius out each time is going to be X. There's our circle. Our height is our highest Y minus our lowest Y. Our highest Y in this case was the 2 -, X ^2. Our lowest Y was the minus X ^2 DX. So if we evaluate this, we have 0 to one 2π R in parentheses. We're going to get two X -, X ^2 -, X ^2 is -2 X squared times An X would be -2 X cubed. So we'd get 2π * X ^2 - 1/2 X to the 4th 0 to one 1 - 1/2 is a half. 1/2 * 2π would give us π cubic units. We look at another one. Y equals square root of XY equals 0 Y equal 2 - X about the X axis. So the first thing we have to do is we have to figure out our bounds. So I'm going to set the two YS equal. This time we're going to have to square each side to get rid of the square root. So X = 4 minus four X + X ^2. Take everything to one side and factor. Now we get two different values for X, and we have to check them against the originals because we squared each side. If I stick in four for X, in the first one, I'd get 4-2, but if I stick 4 into the second one, I'd get four -2 so that point doesn't work. If I stick one and for XI, get one one, and over here I also get one one. So that's the point we want. We're going about the Y axis. So we know that if we're going about the Y axis, we want to have, oh, sorry about the X axis. We know we want to be parallel about. So if we're going about the X axis, we're going to do this in terms of dys. So there's our DY 2π. Our Y values are going to go 0 to one. Our radius here is going to be Y, and our height our furthest to the right minus our furthest to the left. Well, our furthest to the right is going to be the line segment 2 -, X, which if we solve for X, that's really just going to be 2 - y. And then instead of y = sqrt X here, we're going to get X equal y ^2. So the furthest to the right 2 - y minus the furthest to the left y ^2 d Y. When we evaluate this, we're going to get 2π from 0 to one of two y -, y ^2 -, y ^3 d Y, so 2π * y ^2 - 1/3 Y cubed -1 fourth Y to the fourth zero to 1, so 2 Pi 1 -, 1/3 -, 1/4. That would give me twelfths as a common. So 12 twelfths -4 twelfths -3 twelfths would give me 5 twelfths times 2π or five Pi 6 cubic units. Thank you and have a wonderful day.