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# Find the area of the surface obtained by rotating the circle $x^2 + y^2 = r^2$ about the line $y = r$.

## $$4 \pi^{2} r^{2}$$

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Applications of Integration

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were asked to find the area of the serfs obtained by rotating the circle X squared plus y squared equals R squared about the line. Why equals R Well, for the upper semi circle, we have the function F of X is equal to square root of R squared minus X squared. Therefore, it follows that f prime of X. This is going to be negative X over the square root of R squared minus X squared. Now the surface area generated, we'll call it s one. This is the integral from negative are two x equals positive are of two pi times and then in reference to a previous exercise, we have C minus f of X which is the square root of our squared minus X squared times the square root of one plus f prime of X squared or one plus x squared over r squared minus X squared DX. And so we see that we can combine under the radical and get R squared over r squared minus x squared. Also, we see that the function the inte grand, is an even function. So this is the same as two times two pi or four pi times the integral from zero toe are of this is going to be instead of see I mean are here sorry of ar minus the square root of R squared minus X squared times and this becomes the square root of r squared is simply are over the square root of R squared minus X squared. This is because Harrah's positive the X and then we can write this as four pi times the integral from zero to our of and begins R squared over the square root of R squared minus X squared minus are DX and likewise for the lower semi circle we have that f of X is equal to negative square root of R squared minus X squared so that F prime of X is equal to X over the square root of R squared minus X squared. So we have that area of this lower half as to is the integral again from negative arts positive are of two pi times are minus the square root or I guess this is our plus square root of r squared minus X squared times square root of one plus X squared over R squared minus X squared DX We see this is really identical to s one. Except for we have R plus instead of an ar minus squared of R squared minus X squared. So we'll get the same thing as an s one. We get four pi times the integral from zero to our of and here we'll have our squared over the square root of that are squared minus X squared and then we'll have a plus are DX therefore total surface area Yes is going to be s one plus s to which when we have these together we cancel out the our terms on in the instagrams And so we simply get eight pi times the integral from zero r of our squared over the square root of R squared minus X squared T x This could be simplified by taking the anti derivative. So we get eight pi times the inverse sign of X over our evaluated at zero in our and in fact, this is the inverse sign Vex over our times are squared on front And so we get eight pi r squared times inverse sign of one which is pi over two minus in person of zero, which is zero. And so we have four pi squared r squared

Ohio State University

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Applications of Integration

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