What is the volume obtained by rotating the region enclosed by $y = 11 - x$ $y=11-x$ , $y = 3 x + 7$ $y=3x+7$ , and $x = 0$ $x=0$ about the y-axis?

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What is the volume obtained by rotating the region enclosed by $y = 11 - x$ $y=11-x$ , $y = 3 x + 7$ $y=3x+7$ , and $x = 0$ $x=0$ about the y-axis?

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Answer 1 · 2016-03-27T01:36:42

$V = V_{1} - V_{2} = 11, 525 \frac{π}{27}$ $V= V_1-V_2=11,525pi/27$
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Explanation:

Given: $y = 11 - x$ $y=11-x$ , $y = 3 x + 7$ $y=3x+7$ , and $x = 0$ $x=0$
Required: The volume obtained by rotating about the y-axis
Solution strategy: looking at the figure we see that

$V = 2 π [\int_{1}^{11} x (11 - x) d x - \int_{- 2 \frac{1}{3}}^{1} x (3 x + 7) d x]$ $V= 2pi [int_(1)^11x(11-x)dx- int_(-2 1/3)^1x(3x+7)dx]$

$V_{1} = 2 π [{(\frac{11}{2} x^{2} - \frac{x^{3}}{3})}_{1}^{2}]$ $V_1 = 2pi [(11/2x^2-x^3/3)_1^11]$
$V_{2} = 2 π [{(x^{3} + 7 \frac{x^{2}}{2})}_{- 2 \frac{1}{3}}^{3}]$ $V_2 = 2pi [(x^3+7x^2/2)_(-2 1/3)^1]$
$V = V_{1} - V_{2} = 11, 525 \frac{π}{27}$ $V= V_1-V_2=11,525pi/27$

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What is the volume obtained by rotating the region enclosed by $y = 11 - x$ $y=11-x$ , $y = 3 x + 7$ $y=3x+7$ , and $x = 0$ $x=0$ about the y-axis?

1 Answer

Explanation:

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What is the volume obtained by rotating the region enclosed by y=11−xy=11-x, y=3x+7y=3x+7, and x=0x=0 about the y-axis?

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What is the volume obtained by rotating the region enclosed by $y = 11 - x$ $y=11-x$ , $y = 3 x + 7$ $y=3x+7$ , and $x = 0$ $x=0$ about the y-axis?