Volume by Rotating a Curve: Finding the Solid Between Two Curves

mikhailpavel
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Homework Statement


Hey i have a problem here with volume by cylindrical shells. i wanted to find the given volume of the solid obtained by rotating the region bounded by the curves x=1+(y-2)^2 and x=2 about the x axis.


Homework Equations


we tried to integrating 2 phi f(y) dy with upper limit 3 and lower limit 1.


The Attempt at a Solution


we got the answer 33.5 but i don't think it is the correct one.
immediate help will be appreciable.
 
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You are using cylindrical shells, but your formula is incorrect. A typical shell in this problem has a volume of 2pi*y*(x2 - x1)*dy

x2 is the x-value on the line x = 2, and x1 is the x-value on the parabola. You have graphed the region being revolved, right?
 
can u tell me if my upper and lower limits are right because still i am not getting the right answer..i think!
 
Yes, the two curves intersect at (2, 1) and (2, 3) so y ranges between 1 and 3.
 
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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