Setting up an integral for surface area

Abyssnight
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Homework Statement


Set up an integral that represents the area of the surface generated when the region is bounded by x + y2 = 1 and the y-axis, then rotated about the y-axis. All in one-variable.


Homework Equations


SA = 2pi \int xds
possible x2 + y2 = r2

The Attempt at a Solution


I tried to set up an integral as a circle and somehow ended up getting 2pi\int(r/2)(2pi r) dr

When I picture it, it would be a sphere with a radius of 1. Still don't truly understand on how to make a integral to show it
 
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I looked at it again
would one possible answer be: 4pi\int(-y^2 + 1)\sqrt{1 + 4y^2}dy
with limits of integration (0,1)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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