6.2 Examples on Disk Washer Part 4

All right. So, let's look at the same region, but this time I changed the uh rotational axis from y = -3, which which was pretty ugly. So, I change it to um x = to 2, but no matter what, you still need to sketch it.

Okay, hopefully I can sketch this nicely. Okay. Well, something like that.

Okay. And then I'm going to rotate this about this. So, x = to 2.

So, we'll rotate it this way. And that's pretty nice. We're going to have this.

Okay. And the graph is beautiful enough. I believe like this.

Okay. So, this is the region and the cross-section looks like this. Okay.

So, those are the cross-section. and the orientation. So we sum up in this direction.

Okay. In this direction which of course dy okay and if I remember this is two and one. This is one and one.

This is two and eight. Okay. So from here the bounds would be 1 and eight.

Okay. So meaning that the volume is from one. So y from one to eight.

And then we have area dy where area of the cross-section so cross section is a ring so pi r² - r² okay No, it's a disc. Sorry, it's a dis. So, pi r².

Okay, cuz um there's no hole from here. You can see that there's no hole. Okay, and this is the radius R.

Now, here comes the connection. So, this is the radius R. Okay.

and we go all the way back to the original region. And this is R. Whoopsie.

Uh I should say this is R. Okay, which means this is R. Okay, so that's the connection.

Okay. So, r would be um x right - x left. Okay.

And the x right would be the line x = 2 and the right would be y = x cub. And we need to solve it. So this guy x is y to the 1/3.

Okay. And that's my little r. And we're going to do it here.

So pi 2 - y 1/3. Oh, this is not fun. Not fun at all.

Okay. And we'll finish it up. So 1 to 8 pi 2 - y^ 1/3 Well anyway it's still doable even though this is not fun at all equals to pi 1 to 8 multiplied it out I have four I have 4 y 1/3 + y 23 okay and then the anti-derivative would be 4 y - 4 y to the 4 over 3 and then 3 over 4.

So anti-derivative and I'm going to take the power first + one which is 5 over 3 and 3 over 5 evaluate from 1 to 8. So pi 4 * 8 - 3 * 8 4 over 3 + 3 over 5 8^ 5 over 3 and then - 4 - 3 + 3 over 5. So whenever you plug in one that's basically the coefficient.

So let's continue. So this is 32 - 3 * 2 4th 3 over 5 this is 2 5th minus so let's get the answer directly so that's one and then 8 over 5 well tedious calculation 32us This is tough. I don't even know what's that.

Uh 16. So 16 48. Whoa.

Negative. And here comes something tough. 32 minus a over 5.

So just do the calculation carefully. I have -16 here. I have 88 here and this is 80.

Oh, so small. Okay, I have 8 over 5 pi. Okay, so that's the answer for that problem.

Okay, so I'm I'm I'm giving out this I'm giving out this problem, this example compared to number three. It is because I want to show that with the same region. Okay, the different rotational axis actually resulting in different method.

Sometimes it's washer, sometimes disk. Okay, so everything there's no fixed routine. I mean there's no fixed method for the same for any region.

Okay, you have to analyze it. So um sketch the region, sketch the rotation or sketch the solid after rotation. and then try to sketch the um cross-section so that I can draw the connection.

Okay. Drw the connection and build up the formula. Okay.