Shell Method Example: Finding Volume Using Shell Method

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Homework Statement



The area between the y-axis and the line y = x + 1 from y = 2 to
y = 4 is rotated about the x-axis. Use the shell method to find the
area of the resulting solid.



Homework Equations



V = 2pi integral (radius)(height of shell) dx or dy


The Attempt at a Solution



Well I am kind of confused with this problem, it says to find the AREA but isn't the shell method suppose to find the VOLUME?

Here is my attempt:

y= x+1 -----> x=y-1

V = 2pi integral from 2 to 4 y(y-1) dy

I then evaluated it to get = (76/3)pi as the volume.

What did I do wrong?
 
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I'll agree with what you've done so far. The volume is 76*pi/3. If they really do want area I'd add up the area of the parts of the boundary. You've got two cylinders, a disk with a hole in it and a part of a cone. It seems like it shouldn't be too hard, but I don't know what that has to do with the 'shell method'.
 
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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