+0

# 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.

0
234
2
+262

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

#2
+18394
+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

heureka  Feb 17, 2017
Sort:

#1
+75394
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

CPhill  Feb 17, 2017
#2
+18394
+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

heureka  Feb 17, 2017

### 13 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details