SOLUTION FOR THE CONTOUR INTEGRAL PROBLEM Our closed contour now has a curve C that begins at 3 and goes to -3, and becomes complete by virtue of the line L that runs from -3 to 3.

The integral we are asked to find is of the form:

∫_cPdx + Qdy
and we saw in class that such integrals are equal to:

∫∫[(∂Q/∂x) - (∂P/∂y)]dA

In this integral, P = e^x + y²cosx; and Q = e^2y + 2y sinx. Thus, ∂Q/∂x = 2ycos x; and ∂P/∂y = 2ycos x. Therefore, the value of (∂Q/∂x) - (∂P/∂y) is zero and thus the value of the line integral over the entire (and closed) contour C + L is zero.

This means we know that the sum of the line integrals over C and L must sum to zero. We can write this statement as:

∫_{C+L (e^x + y²cos x)dx + (e^2y + 2ysin x)dy = 0}
or
∫_C(e^x + y²cos x)dx + (e^2y + 2ysin x)dy + ∫_L(e^x + y²cos x)dx + (e^2y + 2ysin x)dy = 0

The integral over C is very tough, however, the integral over L is quite straightforward due to the simple (straight line) geometry of L. However, we only have to solve the line integral over L, since we now know the value of the line integral over C will merely be the negative of the line integral over L.

To solve this integral over the line L, set x = t, dx = dt, y = 0 = dy (since our path is entirely along the x axis), and the line integral becomes simply:

∫_-3³ e^tdt
The value of this integral is trivially e³ - e^-3, and so the value of the integral over the upper half of the ellipse C is
e^-3 - e³
[an error occurred while processing this directive]