Integration

Question

Find the volume of the solid generated by revolving the region bounded by the graphs of y equals y = csc(x) and y = square root 2 about the x-axis where 0≤x≤π.

Answer 1

The region is square-root(2) - csc x  on [pi/4, 3*pi/4]

 

Per the table of integrals:

 

sqrt(2)* x - ln| csc x + cot x |  for x in { pi/4, 3*pi/4}

 

3*sqrt(2)*pi/4 - ln | csc*3*pi/4 + cot 3*pi/4 |  - sqrt(2)*pi/4  + ln | csc pi/4 + cot pi/4 |

 

2*sqrt(2) * pi/4 - ln |  sqrt(2) + -1 | - sqrt(2) * pi/4  + ln | sqrt(2) + 1 |

 

2*sqrt(2) * pi/4 - ln | 1 - sqrt(2) | - sqrt(2)*pi/4 + ln | 1 + sqrt(2) |

 

sqrt(2) * pi/4  + ln |  ( (1 +sqrt(2))/(1 - sqrt(2))|

sqrt(2) * pi/4 + ln |   3+ 2*sqrt(2) |  <--- rationalizing the denominator