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Consider the following line integral along the given positively oriented curve. ∫

C

ye
x
dx+2e
x
dy,C is the rectangle with vertices (0,0),(5,0),(5,6), and (0,6) Use Green's theorem to write an equivalent iterated integral. ∫
C

ye
x
dx+2e
2
dy=∫
0

(∫
0

)dydx Use Green's theorem to evaluate the line integral.

1 Answer

2 votes

Answer:

To write an equivalent iterated integral using Green's theorem, we first need to find the curl of the vector field F = (ye^x, 2e^2).

The curl of a vector field F = (P, Q) is given by the formula ∇ x F = ∂Q/∂x - ∂P/∂y, where ∇ is the del operator. In this case, P = ye^x and Q = 2e^2.

Let's find the partial derivatives of Q and P with respect to x and y, respectively:

∂Q/∂x = 0 (since Q doesn't have an x component)

∂P/∂y = e^x (the derivative of ye^x with respect to y is 0, as y is not a function of y)

Now, using the formula for the curl, we have:

∇ x F = (0 - e^x)i + 0j = -e^x i

Next, we need to find the area enclosed by the given rectangle. The rectangle has a width of 5 and a height of 6, so the area is A = 5 * 6 = 30.

Finally, we can use Green's theorem to evaluate the line integral. Green's theorem states that for a vector field F = (P, Q) and a region R bounded by a positively oriented curve C, the line integral of F along C is equal to the double integral of the curl of F over the region R:

∫C ye^x dx + 2e^2 dy = ∬R -e^x dA

Since the region R is a rectangle, we can write the double integral as an iterated integral:

∬R -e^x dA = ∫0^6 ∫0^5 -e^x dx dy

Therefore, the equivalent iterated integral using Green's theorem is:

∫C ye^x dx + 2e^2 dy = ∫0^6 ∫0^5 -e^x dx dy

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