Question

Line Integral; Working with polar coordinates?

How do I calculate the line integral?

C is spiral in polar coordinates of the equation - see attachment for the equations :)

Answer 1

a > 0, b > 0, and phi = 0 and 0 <= phi <= 6pi

Answer 2

I am not even sure where to begin. It looks like the integral is only a part of the first part (dx) and not the dy. I know I can separate the e's if I need to.

Answer 3

\(r = a+b\theta\)

\(0\le \theta \le 6\pi\)

is it ?

Answer 4

Yes

Answer 5

start by parameterizing the curve in terms of \(\theta\) may be :
\(x = r\cos\theta = (a+b\theta)\cos\theta\)
\(y=r\sin\theta =(a+b\theta)\sin\theta\)

Answer 6

Ooh... !

Answer 7

\(x = r\cos\theta = (a+b\theta)\cos\theta\)
\(y=r\sin\theta =(a+b\theta)\sin\theta\)

\(dx = ?\)
\(dy=?\)

Answer 8

What do I do with them?

Answer 9

Asking someone; they said it is.. something gradient? I suspect I have missed a class.

Answer 10

plug the parameterization in the integrand and setup the bounds

Answer 11

replace x and y with ur parameterization in the integrand

Answer 12

What about the dx and dy? Do I insert them as well?

Answer 13

Yes, simply replace them

Answer 14

How do I solve the int... oooo , I will get two dphi's , no?

Answer 15

\(x = r\cos\theta = (a+b\theta)\cos\theta\)
\(y=r\sin\theta =(a+b\theta)\sin\theta\)

\(dx = [b\cos\theta -(a+b\theta)\sin\theta]d\theta\)
\(dy=[b\sin\theta+(a+b\theta)\cos\theta]d\theta\)

\[\int_C2xye^{yx^2}dx+x^2e^{yx^2}dy \\~\\

=\int_0^{6\pi}2(a+b\theta)\cos\theta(a+b\theta)\sin\theta e^{(a+b\theta)\sin\theta((a+b\theta)\cos\theta)^2}[b\cos\theta -(a+b\theta)\sin\theta]d\theta\\+((a+b\theta)\cos\theta)^2e^{(a+b\theta)\sin\theta((a+b\theta)\cos\theta)^2}[b\sin\theta+(a+b\theta)\cos\theta]d\theta
\]

Answer 16

Yeah, an crazy amount of ... everything! Although how do you solve the one integral with two d phi's? Don't you need two integrals for that?

Answer 17

factor out \(d\theta\)

Answer 18

its just one integral, one differential

Answer 19

try simplifying the trig and hope that it gives a nice looking expression in the end :)

Answer 20

Working on it! :)

Answer 21

I don't want to do it, it is a mess lol

Answer 22

It is. I will have to continue this tomorrow.
Its 6AM now. So... later today!

Thanks for the help! :)

Answer 23

np, do let me know if you find any other simpler way to work it... good night :)