# Help setting up integral to find surface area

• Oct 10th 2008, 08:23 PM
Matt164
Help setting up integral to find surface area
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.
• Oct 11th 2008, 05:38 AM
shawsend
For the equation $\displaystyle x=t,;\ y(t)=2\sin(3t)$, the surface areas of the figures around the x and y axes respectively are:

$\displaystyle A_y=2\pi\int x ds$

$\displaystyle A_y=2\pi\int y ds$

with $\displaystyle ds=\sqrt{(x')^2+(y')^2}$

Then for $\displaystyle f(x)=2\sin(3x);\; 0\leq x\leq \pi/2$, in parametric form that's $\displaystyle x(t)=t,\;y(t)=2\sin(3t);\;0\leq t\leq \pi/2$ and thus:

$\displaystyle A_y=2\pi\int_0^{\pi/2} t\sqrt{1+(6\cos(3t))^2}dt$

$\displaystyle A_x=2\pi\int_0^{\pi/2} 2\sin(3t)\sqrt{1+(6\cos(3t))^2}dt$