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?
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Hi
curves y = x and y = x^2 in the first quadrant
{{{graph( 300, 300,-2,2,-2,2,x^2, x) }}}
Graphs intersect at x = 0 and x = 1
A ={{{ int( x , dx, 0, 1)}}} - {{{int(x^2 , dx, 0, 1)}}} |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*int( (f(y))^2 , dy, 0, 1)}}} 
"cup - cone"
V = {{{ pi*int( (y^(1/2))^2 , dy, 0, 1)}}} - {{{ pi*int( y^2 , dy, 0, 1)}}}
Integrate and calculate
V = {{{pi/2 - pi/3 = highlight_green( pi/6) }}}units^3  0r  .52360 units^3

Wish You the Best in your Studies.