- #1
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Question 1)
Integrate y^2 dx + zx dy + dz, along the circle x^2 + y^2 – 2y = 0 to (1, 1, 0)
I am not sure how to begin on this problem. Would it be beneficial to convert to polar coordinates? The thing that is throwing me off is the fact that the circle’s equation is not the normal circle (x^2 + y^2 = r^2), it has that -2y term and it looks like its radius is growing from 0 to sqrt (2)?
Question 2)
Integrate e^x * cos(y) dy – e^x * sin (y) dx,
Over c, which is the broken line from (ln (2), 0), to (0, 1), to (-ln (2), 0) <-- forms a triangle.
Where “e” is the natural exponent.
This is a Green’s Theorem problem I know.
I can identify that,
p(x, y) = -e^x * sin (y)
q(x, y) = e^x * cos (y)
So I should be able to transform the integral into a double integral over some area with the integrant being,
The partial derivative of q with respect to x, minus the partial derivative of p with respect to y, all of that integrated by dx dy.
Well…
The partial of q with respect to x = -e^x * sin (y)
The partial of p with respect to y = -e^x * sin (y)
So subtracting them appropriately, it ends up being the integral of zero dx dy, which is zero itself.
But this cannot be correct (I know this to be wrong).
Integrate y^2 dx + zx dy + dz, along the circle x^2 + y^2 – 2y = 0 to (1, 1, 0)
I am not sure how to begin on this problem. Would it be beneficial to convert to polar coordinates? The thing that is throwing me off is the fact that the circle’s equation is not the normal circle (x^2 + y^2 = r^2), it has that -2y term and it looks like its radius is growing from 0 to sqrt (2)?
Question 2)
Integrate e^x * cos(y) dy – e^x * sin (y) dx,
Over c, which is the broken line from (ln (2), 0), to (0, 1), to (-ln (2), 0) <-- forms a triangle.
Where “e” is the natural exponent.
This is a Green’s Theorem problem I know.
I can identify that,
p(x, y) = -e^x * sin (y)
q(x, y) = e^x * cos (y)
So I should be able to transform the integral into a double integral over some area with the integrant being,
The partial derivative of q with respect to x, minus the partial derivative of p with respect to y, all of that integrated by dx dy.
Well…
The partial of q with respect to x = -e^x * sin (y)
The partial of p with respect to y = -e^x * sin (y)
So subtracting them appropriately, it ends up being the integral of zero dx dy, which is zero itself.
But this cannot be correct (I know this to be wrong).