For the equation , the surface areas of the figures around the x and y axes respectively are:
with
Then for , in parametric form that's and thus:
I need help with the following problem:
Let C be the curve defined by y=2sin(3x) on the interval [0,Pi/2]. I need help setting up the integral required to find the surface area of the surface that results when C is revolved about the specified axis:
a.) The x-axis
b.) The y-axis
I think for a you would use S=2Pi∫(from 0 to Pi/2) 2sin(3x)*sqrt(1+(2cos(3x) *3)^2)
I think for b you would use S=2Pi∫(from 0 to Pi/2) x*sqrt(1+2cos(3x)*3)^2)
Have I set these integrals up correctly or have I done something wrong? Thanks in advance to anyone who can help me.