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

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

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Answer 1 · 2015-09-21T16:10:58

$V = \frac{72 π}{5}$ $V=(72pi)/5$

Explanation:

$x_{1} = y^{2}$ $x_1=y^2$
$x_{2} = y + 2$ $x_2=y+2$

$V = π \int_{y_{1}}^{y_{2}} (x_{2}^{2} - x_{1}^{2}) d y$ $V=pi int_(y_1)^(y_2) (x_2^2-x_1^2)dy$

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

$y_{1} = - 1, y_{2} = 2$ $y_1=-1, y_2=2$

$V = π \int_{- 1}^{2} ({(y + 2)}^{2} - {(y^{2})}^{2}) d y$ $V=pi int_(-1)^(2) ((y+2)^2-(y^2)^2)dy$

$V = π \int_{- 1}^{2} (y^{2} + 4 y + 4 - y^{4}) d y$ $V=pi int_(-1)^(2) (y^2+4y+4-y^4)dy$

$V = π (\frac{y^{3}}{3} + 2 y^{2} + 4 y - \frac{y^{5}}{5}) {∣ ∣ ∣}_{- 1}^{2}$ $V=pi(y^3/3+2y^2+4y-y^5/5)|_-1^2$

$V = π (\frac{8}{3} + 8 + 8 - \frac{32}{5} + \frac{1}{3} - 2 + 4 - \frac{1}{5})$ $V=pi(8/3+8+8-32/5+1/3-2+4-1/5)$

$V = \frac{72 π}{5}$ $V=(72pi)/5$

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

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Explanation:

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How do you find the volume of the solid generated by revolving this region about the y axis, x=y2x= y^2 and x=y+2x= y+2?

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Explanation:

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