Double integral of 4y-2x-4 dA where the region of integration is formed by the v

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It's easier to do with a double integral, but OK, that's not what's asked. Use something similar to the shell method. Take a small slice and rotate around the y-axis, this creates a small ribbon. What's the area of this small cylindrical ribbon? This is the tricky part...especially since I initially did it wrong. Here is the trick. You have a ribbon (that may be on a slant). You essentially have a rectangle...when you unwrap the ribbon it will be base times height: A = bh Well the base is easy enough, it's a circle: b = 2pr The height is a little harder to realize, It's essentially the length of your differential slice: h = ds dA = 2prds But what is ds? To find this you MUST do a pythagorean triangle: ds² = dx² + dy² but dy = (dy/dx)dx = y'dx --> ds² = dx² + y'²dx² = (1 + y'²)dx² --> ds = v(1 + y'²)dx So now we have: dA = 2prv(1 + y'²)dx r is pretty easy too, it's just the distance from the y-axis: x. dA = 2pxv(1 + y'²)dx This is probably the formula you are given. Now do this integral: y' = slope = (5-0)/(3-0) = 5/3 dA = 2pxv(1 + (5/3)²)dx = 2xv((3² + 5²)/3²)dx = v(34)(2/3)pxdx Now the integral is pretty easy, you basically just have: xdx --> x²/2 --> 3²/2 = 9/2 So we have: A = v(34)(2/3)p * 9/2 = 3v34p We can check, because there is a formula for the lateral surface area of a cone: LS = prs r - radius s - slant height s = v(3² + 5²) = v34 So we should get: LS = p * 3 * v34 So we did it right! Edit: I didn't explicitly write down the equation of the line because we don't need the equation we just needed the derivative and since it was a straight line I knew that the derivative was just the slope of the line. Otherwise we would have written the equation of the line: y - 0 = (5/3)(x - 0) --> y = (5/3)x y' = (5/3)

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