# Thread: Surface Area of revolving object

1. ## Surface Area of revolving object

Problem:
The arc of the curve y=arcsec(x) from x=2/sqrt(3) to x=2 is revolved about the line y=pi/2.

Could someone help me set up the integrals to the find surface area by integrating with respect to both x and y? (I could solve myself after I get it set up)

2. Draw yourself the graph and the curve. Now convince yourself the rotating $\sec^{-1}(x)$ around the line $y=\pi/2$ is the same as moving the curve down $\pi/2$ units and rotating about $y=0$. When we move the curve down the new curve is $\sec^{-1} x - \pi / 2$. Thus, find the surface area of revolution of this new curve using the formula that you learned from $x=2/\sqrt{3}$ to $x=2$.

3. Originally Posted by ThePerfectHacker
Draw yourself the graph and the curve. Now convince yourself the rotating $\sec^{-1}(x)$ around the line $y=\pi/2$ is the same as moving the curve down $\pi/2$ units and rotating about $y=0$. When we move the curve down the new curve is $\sec^{-1} x - \pi / 2$. Thus, find the surface area of revolution of this new curve using the formula that you learned from $x=2/\sqrt{3}$ to $x=2$.
Well the formula I have is $S = \int_{c}^{d} 2 * pi * g(x) sqrt{1+g'(x)^2}$ but where does it mean by integrating with respect to x and with respect to y?

4. $2\pi \int_a^b f(x) \sqrt{1+[f'(x)]^2} \ dx$.

As I explain $a=2/\sqrt{3}$ and $b=2$. But instead of $f(x) = \sec^{-1} x$ we have rather $f(x) = \sec^{-1} x - \frac{\pi}{2}$.

Now continue.