Asked by Beth
Compute the volume of the solid formed by revolving the fourth quadrant region bounded by y = x^2 - 1 , y = 0, and x = 0 about the line y = 6.
Answers
Answered by
Steve
If you use discs, each disc has a hole in it or radius 6. So, the volume is
Int(pi*(R^2-r^2) dx)[0,1]
= Int(pi*(7-y)^2 - 7^2)dx)[0,1]
= 148/15 pi
If you use shells, each shell has height x, thickness dy.
x = (y+1)^1/2
Int(2pi*(7-y)*x dy)[-1,0]
2pi*Int(7-y)*sqrt(y+1) dy)[-1,0]
= 148/15 pi
Int(pi*(R^2-r^2) dx)[0,1]
= Int(pi*(7-y)^2 - 7^2)dx)[0,1]
= 148/15 pi
If you use shells, each shell has height x, thickness dy.
x = (y+1)^1/2
Int(2pi*(7-y)*x dy)[-1,0]
2pi*Int(7-y)*sqrt(y+1) dy)[-1,0]
= 148/15 pi
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.