How did you get 2-6y for the double integral?
A task says:
Evaluate I = "closed integral"C (sinx + 3y^2)dx + (2x - e^((-y)^2))dy, where C is the boundary of the half-disk x^2 + y^2 <_ a^2, y >_ 0, oriented counterclockwise.
When you use Green's Theorem on I you get a "double integral"(2 - 6y) dyda. I am insecure about choosing the right limits for x and y. My purpose is:
x has an upper limit "pi", lower limit a^2. y has an upper limit "pi" and it's lower limit is a^2. Is this q correct setup?
Excuse my poor notation, I just signed up,
The "2- 6y" is the correct integrand. As for the limits of integration, " " would apply if you were integrating in polar coordinates!
To find the limits of integration in x, y coordinates, start by drawing a picture. " , is a semicircle with base in the x axis from x= -a to x= a. For each x y runs from 0 up to the semicircle: (from solving for y).
You could integrate in the opposite order: Overall, y goes from 0 to a and, for each y, x goes from on the left to
Or, if you really wanted to (I don't recommend it) you could integrate in polar coordinates: the angle goes from 0 to and r goes from 0 to a, independently. The "hard part" is that you need to convert the integrand into polar coordinates. . And, of course, the "differential of area" in polar coordinates is
The integrand is the divergence of the F vector:
div F = (2 - 0) - (0 + 6y) = 2 - 6y
The first part is derived with respect to x, thus the exponential function dies. The second part is differentiated with respect to y, hence sin(x) = 0.