1. Evaluate the following line integral:
∫C ydx - 2xdy , where (C) is the portion of y = x^2 from (2,4) to (0,0)
I applied Green's Theorem but I am not sure if it is the right way to solve it.[/quote[
Let x= t,
y=t2 Then dx= dt and dy= 2tdt. As (x, y) goes from (2, 4) to (0, 0), t goes from 2 to 0. The integral is
∫20t2dt−(2t)(2tdt)=∫20(t2−4t2)dt=−3∫20t2dt=3∫02t2dt.
2. . Evaluate the triple integral:
∭Q 2y + z dV
Where Q is the solid bounded by z = 6 - x^2 - y^2 and z = 2.
z=6−x2−y2 is a paraboloid, opening downward with vertex at (0, 0, 6). It cuts the plane z= 2 where
2=6−x2−y2 so where
x2+y2=6−2=4. That is the circle with center at (0, 0, 2) and radius 2.
I would use cylindrical coordinates. Since the "base" at z= 2 is the full circle,
θ goes from 0 to
2π. Since the radius of that circle is 2, r goes from 0 to 2. And, of course, z goes from 2 to
6−x2−y2=6−r2.
As a triple integral that is
∫02π∫02∫06−r2rdzdrdθ.