(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The region R enclosed by the curves x-2y=-2 and y=sqrt(x-2)+2 is rotated about the line x=-2. Find the volume of the resulting solid.

2. Relevant equations

[tex]$\displaystyle \Large V=\int _b^d A(y) dy$[/tex]

[tex]$\displaystyle \Large V=\int _b^d\pi ((2 +x_R(y))^2-(2 +x_L(y))^2)dy.$[/tex]

3. The attempt at a solution

after making y_{1}= y_{2}

intersection points, (2, 2) and (6, 4).

i need x_{R}and x_{L}, so

x_{R}= 2y-2

x_{L}= y^{2}-4y+6

[tex]$\displaystyle \Large V=\int _2^4\pi [(2 +(2y-2))^2-(2+(y^2-4y+6))^2]dy$[/tex]

==>

[tex]$\displaystyle \Large V=\int _2^4\pi [4y^2-(y^4 - 8y^3 +32y^2 +64y +64)]dy$[/tex]

==>

[tex]$\displaystyle \Large V=\pi [(-y^5)/5 + 2y^4 -(28y^3)/3 +32y -64y)]$[/tex]^{4}_{2}

V = 261.3805088

what's wrong with my calculation?

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# Homework Help: Volume Around x=-r

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