5. Let F⃗ = y cos x i⃗ + (sin x + y)j⃗ a) Find a potential...

5. Let \(\vec{F} = y \cos x \vec{i} + (\sin x + y)\vec{j}\) a) Find a potential function for \(\vec{F}\), that is a function f such that \(\vec{F} = \nabla f\) b) Without any parameterization of the curve C, compute \(\int_C \vec{F} \cdot d\vec{r}\) where C is the upper half of the circle of center (2,0) and radius 4, traversed in the clockwise direction. 6. Evaluate \(\int_C (1 + \tan x)dx + (x^2 + e^y)dy\) given that C is the positively oriented boundary of the region enclosed by the curves \(y = \sqrt{x}\), \(x = 1\) and \(y = 0\)

Transcript

00:01 In this problem we are given that f equals to 2y plus e raised to the power of x i vector plus x plus sign of y squared times j vector.

00:18 And we are provided with the curve c which is given by x squared plus y squared equals to 1 which is the unit circle.

00:26 And we are asked to evaluate the integral of fds over the curves.

00:32 By making use of the greens theorem, we know that the integral over the curve c of p dx plus q dy equals to the double integral of the region d enclosed by the curve of del q over del x minus del p over del y where del denotes the partial derivative dx, d .y.

01:02 So first, let us compare the green theorem with the given field in that case we have p to be equal to 2y plus e raise to the power of x and q equals to x plus sign of y squared let us partially differentiate p with respect to y we get two and partially differentiating q with respect to x we get one now let us subtract the partial derivative of q with respect to x and the partial derivative of p with respect to y we get 1.

01:35 Minus 2 which equals to negative 1.

01:38 So now let us convert the entire question in terms of polar polar coordinates.

01:45 That is we have x to be equal to r times cost of theta and y equals to r times sine of theta and here since we have a unit circle r varies from 0 to 1 and teta varies from 0 to 2 pi and clearly here we would have d x, dy to be equal to r times dr d -teta.

02:08 So substituting all of the values in the formula of the greens theorem, we have double integral, integral 0 to 2 -pi, integral 0 to 1, negative 1 times r -dr -d -teta...

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