What is the volume of the solid produced by revolving $f (x) = cot x, x \in [\frac{π}{4}, \frac{π}{2}]$ $f(x)=cotx, x in [pi/4,pi/2]$ around the x-axis?

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What is the volume of the solid produced by revolving $f (x) = cot x, x \in [\frac{π}{4}, \frac{π}{2}]$ $f(x)=cotx, x in [pi/4,pi/2]$ around the x-axis?

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Answer 1 · 2018-03-09T11:10:12

$V = π - \frac{1}{4} π^{2}$ $V=pi-1/4pi^2$

Explanation:

The formula for finding the volume of a solid produced by revolving a function $f$ $f$ around the $x$ $x$ -axis is

$V = \int_{a}^{b} π {[f (x)]}^{2} d x$ $V=int_a^bpi[f(x)]^2dx$

So for $f (x) = cot x$ $f(x)=cotx$ , the volume of its solid of revolution between $π / 4$ $pi"/"4$ and $π / 2$ $pi"/"2$ is

$V = \int_{π / 4}^{π / 2} π {(cot x)}^{2} d x = π \int_{π / 4}^{π / 2} {cot}^{2} x d x = π \int_{π / 4}^{π / 2} {csc}^{2} x - 1 d x = - π {[cot x + x]}_{π / 4}^{π / 2} = - π ((0 - 1) + (\frac{π}{2} - \frac{π}{4})) = π - \frac{1}{4} π^{2}$ $V=int_(pi"/"4)^(pi"/"2)pi(cotx)^2dx=piint_(pi"/"4)^(pi"/"2)cot^2xdx=piint_(pi"/"4)^(pi"/"2)csc^2x-1dx=-pi[cotx+x]_(pi"/"4)^(pi"/"2)=-pi((0-1)+(pi/2-pi/4))=pi-1/4pi^2$

Answer 2 · 2018-03-09T11:19:46

$Area of revolution around$ $"Area of revolution around"$ $x -axis = 0.674$ $x"-axis"=0.674$

Explanation:

$Area of revolution around$ $"Area of revolution around"$ $x -axis = π \int_{a}^{b} {(f (x))}^{2} d x$ $x"-axis"=piint_a^b(f(x))^2dx$

$f (x) = cot x$ $f(x)=cotx$
${f (x)}^{2} = cot x$ $f(x)^2=cotx$

$\int_{\frac{π}{4}}^{\frac{π}{2}} {cot}^{2} x d x = \int_{\frac{π}{4}}^{\frac{π}{2}} {csc}^{2} x - 1 d x$ $int_(pi/4)^(pi/2)cot^2xdx=int_(pi/4)^(pi/2)csc^2x-1dx$
$\int_{\frac{π}{4}}^{\frac{π}{2}} {cot}^{2} x d x = π {[- cot x - x]}_{\frac{π}{4}}^{\frac{π}{2}}$ $color(white)(int_(pi/4)^(pi/2)cot^2xdx)=pi[-cotx-x]_(pi/4)^(pi/2)$
$\int_{\frac{π}{4}}^{\frac{π}{2}} {cot}^{2} x d x = π [(- cot (\frac{π}{2}) - \frac{π}{2}) - (- cot (\frac{π}{4}) - \frac{π}{4})]$ $color(white)(int_(pi/4)^(pi/2)cot^2xdx)=pi[(-cot(pi/2)-pi/2)-(-cot(pi/4)-pi/4)]$
$\int_{\frac{π}{4}}^{\frac{π}{2}} {cot}^{2} x d x = π [(- 0 - \frac{π}{2}) - (- 1 - \frac{π}{4})]$ $color(white)(int_(pi/4)^(pi/2)cot^2xdx)=pi[(-0-pi/2)-(-1-pi/4)]$
$\int_{\frac{π}{4}}^{\frac{π}{2}} {cot}^{2} x d x = π [- \frac{π}{2} + 1 + \frac{π}{4}]$ $color(white)(int_(pi/4)^(pi/2)cot^2xdx)=pi[-pi/2+1+pi/4]$
$\int_{\frac{π}{4}}^{\frac{π}{2}} {cot}^{2} x d x = 0.674$ $color(white)(int_(pi/4)^(pi/2)cot^2xdx)=0.674$

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What is the volume of the solid produced by revolving $f (x) = cot x, x \in [\frac{π}{4}, \frac{π}{2}]$ $f(x)=cotx, x in [pi/4,pi/2]$ around the x-axis?

2 Answers

Explanation:

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What is the volume of the solid produced by revolving f(x)=cotx,x∈[π4,π2]f(x)=cotx, x in [pi/4,pi/2] around the x-axis?

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Explanation:

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What is the volume of the solid produced by revolving $f (x) = cot x, x \in [\frac{π}{4}, \frac{π}{2}]$ $f(x)=cotx, x in [pi/4,pi/2]$ around the x-axis?