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The region bounded by y = 5 and y = x+(4/x) is rotated about the line x=−1. Find the volume of the resulting solid by any method.

gretzu  Feb 17, 2017

Best Answer 

 #2
avatar+19632 
+15

The region bounded by y = 5 and y = x+(4/x) is rotated about the line x=−1.

Find the volume of the resulting solid by any method.

 

Shell method:

\(\begin{array}{|rcll|} \hline f\left(x\right) &=& x+\frac{4}{x} \\ g\left(x\right) &=& 5 \\ \hline \end{array}\)

 

Rotation around Vertical line: \(x_\text{axis of rotation}=-1\)

Radius of Rotation: \(r = x-x_\text{axis of rotation} =x-(-1)=x+1\)

Circumference of the cylinder \(= 2\pi r\)

Shell Volume \(= 2\pi r \cdot[ g(x)-f(x)]\ dx\)

 

Limits of integration:

\(\begin{array}{|rcll|} \hline f\left(x\right) &=& g\left(x\right) \\ x+\frac{4}{x} &=& 5 \quad & | \quad \cdot x \\ x^2+4 &=& 5x \quad & | \quad -5x \\ x^2- 5x +4 &=& 0 \\ (x-1)(x-4) &=& 0 \\ a = 1 &\text{and}& b = 4\\ \hline \end{array} \)

 

Volume:

\(\begin{array}{|rcll|} \hline V &=& \int \limits_{a}^{b} 2\pi r \cdot[ g(x)-f(x)] \ dx \\ &=& 2\pi \cdot \int \limits_{a}^{b} r \cdot[ g(x)-f(x)] \ dx \\ \hline \end{array} \)

\(\small{ \begin{array}{|rcll|} \hline V &=& 2\pi \cdot \int \limits_{1}^{4} (x+1) \cdot \left[ 5-(x+\frac{4}{x}) \right] \ dx \\ &=& 2\pi \cdot \int \limits_{1}^{4} [ x\cdot( 5-x-\frac{4}{x}) + 5-x-\frac{4}{x} ] \ dx \\ &=& 2\pi \cdot \int \limits_{1}^{4} ( 5x-x^2-4 + 5-x-\frac{4}{x} ) \ dx \\ &=& 2\pi \cdot \int \limits_{1}^{4} ( 4x-x^2+1-\frac{4}{x} ) \ dx \\ &=& 2\pi \cdot [~ 2x^2-\frac{x^3}{3}+x-4\cdot \ln(x) ~]_1^4 \\ &=& 2\pi \cdot \left[~ 2\cdot (4)^2-\frac{(4)^3}{3}+ (4) -4\cdot \ln(4) - \left(2\cdot (1)^2-\frac{(1)^3}{3}+(1)-4\cdot \ln(1) \right) \right]\\ &=& 2\pi \cdot \left[32-\frac{64}{3}+4-4\cdot \ln(4) - \left(2-\frac{1}{3}+1+0 \right) \right] \\ &=& 2\pi \cdot \left[32-\frac{64}{3}+4-4\cdot \ln(4) - 2+\frac{1}{3}-1 \right] \\ &=& 2\pi \cdot \left[33-\frac{63}{3}-4\cdot \ln(4) \right] \\ &=& 2\pi \cdot \left[33-21-4\cdot \ln(2^2) \right] \\ &=& 2\pi \cdot \left[12-4\cdot 2\cdot \ln(2) \right] \\ &=& 2\pi \cdot [12-8\cdot \ln(2) ] \\ &=& 2\pi \cdot 6.45482255552 \\ &=& 40.5568462413 \\ \hline \end{array} } \)

 

 

The volume of the resulting solid is 40.5568462413

 

laugh

heureka  Feb 17, 2017
 #1
avatar+87301 
0

The shell method seems logical here.....we have

 

      4

2 pi ∫  (x + 1)[ 5 - (x + 4/x)]  dx  =

      1

 

      4

2 pi ∫   5x - x^2 - 4 + 5 - x - 4/x     dx

      1

 

 

       4

2 pi ∫   4x - x^2 + 1- 4/x     dx    =  [16137 / 1250]  pi units^3

      1

 

 

 

cool cool cool

CPhill  Feb 17, 2017
 #2
avatar+19632 
+15
Best Answer

The region bounded by y = 5 and y = x+(4/x) is rotated about the line x=−1.

Find the volume of the resulting solid by any method.

 

Shell method:

\(\begin{array}{|rcll|} \hline f\left(x\right) &=& x+\frac{4}{x} \\ g\left(x\right) &=& 5 \\ \hline \end{array}\)

 

Rotation around Vertical line: \(x_\text{axis of rotation}=-1\)

Radius of Rotation: \(r = x-x_\text{axis of rotation} =x-(-1)=x+1\)

Circumference of the cylinder \(= 2\pi r\)

Shell Volume \(= 2\pi r \cdot[ g(x)-f(x)]\ dx\)

 

Limits of integration:

\(\begin{array}{|rcll|} \hline f\left(x\right) &=& g\left(x\right) \\ x+\frac{4}{x} &=& 5 \quad & | \quad \cdot x \\ x^2+4 &=& 5x \quad & | \quad -5x \\ x^2- 5x +4 &=& 0 \\ (x-1)(x-4) &=& 0 \\ a = 1 &\text{and}& b = 4\\ \hline \end{array} \)

 

Volume:

\(\begin{array}{|rcll|} \hline V &=& \int \limits_{a}^{b} 2\pi r \cdot[ g(x)-f(x)] \ dx \\ &=& 2\pi \cdot \int \limits_{a}^{b} r \cdot[ g(x)-f(x)] \ dx \\ \hline \end{array} \)

\(\small{ \begin{array}{|rcll|} \hline V &=& 2\pi \cdot \int \limits_{1}^{4} (x+1) \cdot \left[ 5-(x+\frac{4}{x}) \right] \ dx \\ &=& 2\pi \cdot \int \limits_{1}^{4} [ x\cdot( 5-x-\frac{4}{x}) + 5-x-\frac{4}{x} ] \ dx \\ &=& 2\pi \cdot \int \limits_{1}^{4} ( 5x-x^2-4 + 5-x-\frac{4}{x} ) \ dx \\ &=& 2\pi \cdot \int \limits_{1}^{4} ( 4x-x^2+1-\frac{4}{x} ) \ dx \\ &=& 2\pi \cdot [~ 2x^2-\frac{x^3}{3}+x-4\cdot \ln(x) ~]_1^4 \\ &=& 2\pi \cdot \left[~ 2\cdot (4)^2-\frac{(4)^3}{3}+ (4) -4\cdot \ln(4) - \left(2\cdot (1)^2-\frac{(1)^3}{3}+(1)-4\cdot \ln(1) \right) \right]\\ &=& 2\pi \cdot \left[32-\frac{64}{3}+4-4\cdot \ln(4) - \left(2-\frac{1}{3}+1+0 \right) \right] \\ &=& 2\pi \cdot \left[32-\frac{64}{3}+4-4\cdot \ln(4) - 2+\frac{1}{3}-1 \right] \\ &=& 2\pi \cdot \left[33-\frac{63}{3}-4\cdot \ln(4) \right] \\ &=& 2\pi \cdot \left[33-21-4\cdot \ln(2^2) \right] \\ &=& 2\pi \cdot \left[12-4\cdot 2\cdot \ln(2) \right] \\ &=& 2\pi \cdot [12-8\cdot \ln(2) ] \\ &=& 2\pi \cdot 6.45482255552 \\ &=& 40.5568462413 \\ \hline \end{array} } \)

 

 

The volume of the resulting solid is 40.5568462413

 

laugh

heureka  Feb 17, 2017

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