How do you find the volume of the solid generated by revolving this region about the y axis, $x= y^2$ and $x= y+2$ ?

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Answer 1 · 2015-09-21T16:10:58

$V=(72pi)/5$

$x_1=y^2$
$x_2=y+2$

$V=pi int_(y_1)^(y_2) (x_2^2-x_1^2)dy$

$y_1,y_2=?$
$x_1=x_2$
$y^2=y+2$
$y^2-y-2=0$
$y^2-y-2=y^2-2y+y-2=y(y-2)+y-2=(y-2)(y+1)=0$

$y_1=-1, y_2=2$

$V=pi int_(-1)^(2) ((y+2)^2-(y^2)^2)dy$

$V=pi int_(-1)^(2) (y^2+4y+4-y^4)dy$

$V=pi(y^3/3+2y^2+4y-y^5/5)|_-1^2$

$V=pi(8/3+8+8-32/5+1/3-2+4-1/5)$

$V=(72pi)/5$

How do you find the volume of the solid generated by revolving this region about the y axis, x= y^2x= y^2 and x= y+2x= y+2?