## 6.16850

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Applications of Integration

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Okay. So for this problem, we are asked to set up a integral for the volume where it's rotating about the boundaries provided and then calculate the integral up to five decimal places. So we have that. Why equals zero. Why equals co sign Squared X and then we have between negative pi over two at, uh and pi over two for X. So the first part we want to do is across the x axis. So what we want to do is we want to set up our, um are too great. And so are we going to use the disk method for this? And so what we want to do is we want to set up our volume thio equal pi along the integral beat from B to A or A to B R squared DX. So what I want to do is my are is actually going to be my co sign Squared X. So what I'm going to do is it's going to be pie from pi over two to negative pi over two for are integral and then cosine squared X squared D X. And then, of course, this is going to be pie from pi over two to negative pi over two and then it's gonna be cosign to the fourth x dx. So what I can also do since it's going to be symmetrical because it's a cosine value. What we want to do is go ahead and set this up to be two pi and then either choose negative pi over 2 to 0 or zero to pi over two. I prefer dealing with positives, so co sign for X dx and then by utilizing a calculator, we can find out that this is going to be three pi squared over eight, which is approximately 3.70110 And then, of course, for part B were asked to do this across. Um, it's why equals one. And so what we want to do is use the very we only use a washer method for this one. So just a zey refresher. The washer method is volume equals pie from a to B of our one squared minus are two squared dx. And so since we're now dealing with y equals one, that's gonna be our our one. So we're gonna have pie from, uh, oops from, uh, pi over two to negative pi over two, and we're gonna have one squared minus cosine squared, X squared DX. So what we're going to do for this is it's gonna be pi from negative to positive pi over two. And now we're going to square this, so it's going to end up being one minus and then we I'm sorry, this is actually supposed to be one minus cosine squared X because if we think about the graph, we're gonna have our cosign, and then we're gonna have one. So our coastlines kind of kind of look something like that. So we're gonna have to have that difference, but from one to co sign squared, and then we're gonna have one minus that. So we're gonna have one minus two cosine squared X plus cosign to the fourth ex D X. And then, of course, are ones air going to cancel. So we're gonna have one minus one, which will be zero, and then we have negative. So it could be to cosign squared X minus co signed to the fourth Ex D X. And then So then again, we want to do something similar to what we did with the last one and do two pi from pi over 2 to 0 of to cosign squared X minus cosine to the fourth ex DX. And then if we utilize a calculator, it will tell us that the answer is five. Pi squared over eight, which is approximately 6.16850

#### Topics

Applications of Integration

Lectures

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