SOLUTION: The region bounded by curves y = x and y = x^2 in the first quadrant of the xy-plane is rotated about the y-axis. What is the area and volume of the resulting solid of revolution?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The region bounded by curves y = x and y = x^2 in the first quadrant of the xy-plane is rotated about the y-axis. What is the area and volume of the resulting solid of revolution?      Log On


   

Question 1172395: The region bounded by curves y = x and y = x^2 in the first quadrant of the
xy-plane is rotated about the y-axis. What is the area and volume of the resulting solid of revolution?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
The region bounded by curves y = x and y = x^2 in the first quadrant of the
xy-plane is rotated about the y-axis. What is the area and volume of the resulting solid of revolution?


Hi

curves y = x and y = x^2 in the first quadrant

graph%28+300%2C+300%2C-2%2C2%2C-2%2C2%2Cx%5E2%2C+x%29+

Graphs intersect at x = 0 and x = 1

A =+int%28+x+%2C+dx%2C+0%2C+1%29 - int%28x%5E2+%2C+dx%2C+0%2C+1%29 |top curve - bottom curve

Integrate

A = x^2/2 - x^3/3 from 0 to 1

A = 1/2-1/3 = 1/6 units^2



 volume of the resulting solid of revolution rotated about y-axis

curves x = y and x = y^(1/2) in the first quadrant

V = =+pi%2Aint%28+%28f%28y%29%29%5E2+%2C+dy%2C+0%2C+1%29 

"cup - cone"

V = +pi%2Aint%28+%28y%5E%281%2F2%29%29%5E2+%2C+dy%2C+0%2C+1%29 - +pi%2Aint%28+y%5E2+%2C+dy%2C+0%2C+1%29

Integrate and calculate

V = pi%2F2+-+pi%2F3+=+highlight_green%28+pi%2F6%29+units^3  0r  .52360 units^3



Wish You the Best in your Studies.