v = ∫∫R f(x,y) dA
= ∫[0,1]∫[x,x^2] 2x^2 y dy dx
= ∫[0,1] (x^2 y^2)[x,x^2] dx
= ∫[0,1] (x^6 - x^4) dx
= (1/7 x^7 - 1/5 x^5)[0,1]
= 1/7 - 1/5
= -2/35
find the volume of the solid bounded above by the surface z=f(x,y) and below by the plane region R
f(x,y)=2x^2 y;R is the region bounded by the graphs of y=x and y=x^2
2 answers
Oops. That sign is reversed, since y f goes from x^2 to x.
Answer is 2/35
Couldn't figure how I got a minus value for a function that is positive over the region.
Answer is 2/35
Couldn't figure how I got a minus value for a function that is positive over the region.