How do we evaluate the volume under the surface given by, z=2xsin(y) over the region bounded below by the curve y=x^2 and above by the line y=0 for 0<=x<=1
Can anyone please clarify my doubys regarding the limits of this integral?
Limits of,
x : 0-1(as it is given "above the line y=0 for 0<=x<=1
Am i right?)
y : 0- x^2 ( as it is given "below by the curve y=x^2"
Am i right?)
Can you please explain how to get the limits of z?
I'm a bit unclear on what you mean by
bounded .. above by the line y=0
y=0 is the x-z plane, which would not bound it above.
Since x goes from 0 to 1, then y would also go from 0 to 1.
z goes from 0 to 2x siny, so my feeling is that you want
v = ∫[0,1] ∫[x^2,1] ∫[0,2x siny] dz dy dx
= ∫[0,1] ∫[x^2,1] 2x siny dy dx
= ∫[0,1] -2x cosy dx [x^2,1]
= ∫[0,1] -2x cos(1) + 2x cos(x^2) dx
= -cos(1) x^2 + sin(x^2) [0,1]
= sin(1) - cos(1)
If I got the boundaries wrong, maybe you can fix them and follow my steps.