How do you find the volume of the solid obtained by rotating the region bounded by the curves $y= x^2 - 4$ and $y= 3x$ and $x=0$ about the y axis?

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Answer 1 · 2015-09-28T02:54:58

$136pi$

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Let's find intersection:

$x=y/3 => y=(y/3)^2-4$

$y^2/9-y-4=0$

$y^2-9y-36=0$
$y^2-12y+3y-36=y(y-12)+3(y-12)=0$
$(y-12)(y+3)=0 <=> y=12 vv y=-3$

$y=x^2-4 => x=sqrt(y+4)$
$y=3x => x=y/3$

$V=pi (int_-4^12 (sqrt(y+4))^2dy - int_0^12 (y/3)^2dy)$

$V=pi (int_-4^12 (y+4)dy - int_0^12 y^2/9dy)$

$V_1=pi [y^2/2+4y]|_-4^12$

$V_1=pi (144+48-8+16)=200pi$

$V_2=pi [y^3/27]|_0^12$

$V_2=64pi$

$V=200pi-64pi = 136pi$

How do you find the volume of the solid obtained by rotating the region bounded by the curves y= x^2 - 4 y= x^2 - 4 and y= 3xy= 3x and x=0x=0 about the y axis?