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 $y = 3 x$ $y= 3x$ and $x = 0$ $x=0$ about the y axis?

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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 $y = 3 x$ $y= 3x$ and $x = 0$ $x=0$ about the y axis?

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Answer 1 · 2015-09-28T02:54:58

$136 π$ $136pi$

Explanation:

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Let's find intersection:

$x = \frac{y}{3} \Rightarrow y = {(\frac{y}{3})}^{2} - 4$ $x=y/3 => y=(y/3)^2-4$

$\frac{y^{2}}{9} - y - 4 = 0$ $y^2/9-y-4=0$

$y^{2} - 9 y - 36 = 0$ $y^2-9y-36=0$
$y^{2} - 12 y + 3 y - 36 = y (y - 12) + 3 (y - 12) = 0$ $y^2-12y+3y-36=y(y-12)+3(y-12)=0$
$(y - 12) (y + 3) = 0 \Leftrightarrow y = 12 \lor y = - 3$ $(y-12)(y+3)=0 <=> y=12 vv y=-3$

$y = x^{2} - 4 \Rightarrow x = \sqrt{y + 4}$ $y=x^2-4 => x=sqrt(y+4)$
$y = 3 x \Rightarrow x = \frac{y}{3}$ $y=3x => x=y/3$

$V = π (\int_{- 4}^{12} {(\sqrt{y + 4})}^{2} d y - \int_{0}^{12} {(\frac{y}{3})}^{2} d y)$ $V=pi (int_-4^12 (sqrt(y+4))^2dy - int_0^12 (y/3)^2dy)$

$V = π (\int_{- 4}^{12} (y + 4) d y - \int_{0}^{12} \frac{y^{2}}{9} d y)$ $V=pi (int_-4^12 (y+4)dy - int_0^12 y^2/9dy)$

$V_{1} = π [\frac{y^{2}}{2} + 4 y] {∣ ∣ ∣}_{- 4}^{12}$ $V_1=pi [y^2/2+4y]|_-4^12$

$V_{1} = π (144 + 48 - 8 + 16) = 200 π$ $V_1=pi (144+48-8+16)=200pi$

$V_{2} = π [\frac{y^{3}}{27}] {∣ ∣ ∣}_{0}^{12}$ $V_2=pi [y^3/27]|_0^12$

$V_{2} = 64 π$ $V_2=64pi$

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

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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 $y = 3 x$ $y= 3x$ and $x = 0$ $x=0$ about the y axis?

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How do you find the volume of the solid obtained by rotating the region bounded by the curves y=x2−4y= x^2 - 4 and y=3xy= 3x and x=0x=0 about the y axis?

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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 $y = 3 x$ $y= 3x$ and $x = 0$ $x=0$ about the y axis?