The region under the curves $y=cosx-sinx, 0<=x<=pi/4$ is rotated about the x axis. How do you sketch the region and find the volumes of the two solids of revolution?

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Answer 1 · 2018-02-22T03:49:22

volume = integration of[(pi*((cosx - sinx)^2)dx] from x = 0 to x = pi/4

volume = integration of[(pi*(1- sin2x)dx] from x = 0 to x = pi/4

volume = [(pi*(x + cos2x/2)] from x = 0 to x = pi/4

volume = [(pi(pi/4 + cospi/2)] -[(pi(0/4 + cos0/2)]
volume = [(pi(pi/4 +0)] -[(pi(0 + 1)]
volume =pi^2/4 -pi

The region under the curves y=cosx-sinx, 0<=x<=pi/4y=cosx-sinx, 0<=x<=pi/4 is rotated about the x axis. How do you sketch the region and find the volumes of the two solids of revolution?