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, \(\displaystyle y= t^2\) 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 \(\displaystyle \int_2^0 t^2dt- (2t)(2tdt)= \int_2^0 (t^2- 4t^2)dt= -3\int_2^0 t^2dt= 3\int_0^2 t^2dt\).

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.

\(\displaystyle z= 6- x^2- y^2\) is a paraboloid, opening downward with vertex at (0, 0, 6). It cuts the plane z= 2 where \(\displaystyle 2= 6- x^2- y^2\) so where \(\displaystyle x^2+ y^2= 6- 2= 4\). 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, \(\displaystyle \theta\) goes from 0 to \(\displaystyle 2\pi\). Since the radius of that circle is 2, r goes from 0 to 2. And, of course, z goes from 2 to \(\displaystyle 6- x^2- y^2= 6- r^2\).

As a triple integral that is \(\displaystyle \int_0^{2\pi}\int_0^2\int_0^{6- r^2} r dzdrd\theta\).