Area by washers and/or cylindrical shells: other than x,y

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What if you want to rotate around something other than the x/y axis?

For example:

Homework Statement



y=x, y=0, x=1, rotated around the line x=-1

Homework Equations



NumberedEquation1.gif


or

NumberedEquation1.gif


The Attempt at a Solution



V= ⌠(between 0 and 1)π[1+x]^2 dx

= π(1/3(x)^3+x^2+2x),x=0, x=1

=π((1/3)(1)^3+(1)^2+2(1))-0

= 10π/3

Shell method:

V= ⌠(between 0 and 1)2πx(x+1)dx

=2π((1/2)x^2+(1/3)x^3)x=1,x=0

=2π((1/2)+(1/3))-0

=2π(5/6)

=5π/3

Obviously the two answers do not match (i don't even know if i am correct on either); where am i going wrong?

Thanks in advance!
 
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unless I am mistaken, if you're using the first equation you have to do it with respect to y...

and when you did it the second way, the expression inside of the integral is wrong.
 
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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