Use the method of cylindrical shells to find the volume of the solid generated by revolving aabout the indicated axis the region bounded by the given curves.(adsbygoogle = window.adsbygoogle || []).push({});

x=y, x+2y=3, y=0; about the x axis.

Latest version of my work:

(I don't know how to paste a graph into the post. But I graphed my curves.)

[tex]V=\int_{a}^{b}2\pi\L(x)dx[/tex]

But since this rotates around the x axis instead of the y axis, replace all x's with y's.

I took the 2pi out of the integral b/c it's a constant. After that I'm very unsure of what's going on.

y stayed y.

L(x), the height of the cylinder, is now the width of the cylinder, and is defined as the length of the horizontal line, which is the distance between the two curves.

So L(x)= (-2y+3) - y = -3y+3.

My interval is the largest distance beteween my two curves, which is the x-axis on the interval [0,3]

I shoved that into the equation, factored out a 3 and removed it from the integrand, leaving me with

[tex]6\pi\int_{0}^{3}(y-3y)dx\\=(6\pi)[\frac{y^2}{2}-\frac{y^3}{3}]^3_0\\=6\pi\frac{-27}{6}\\=-27\pi[/tex]

But the book says the answer is pi.

Help? Please!!

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# Cylindrical shell

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