Line Integration

**purplerain** · Nov 21st 2009, 08:36 AM

$\int _c xy^4ds$ C is the right half of the circle $x^2+y^2=16$

Here is what i did. Since C is the right half of the circle my bounds are from $-\pi/2 \leq t \leq \pi/2$ . I got x = 4cost and y = 4sint. I then take the arclength of that, and multiple that by the original function with the polar coordinates substituted into it which is $\int (4cost)(4sint)^4\sqrt{(-4sint^2)+(4cost^2)}dt$

Im kind of confused on how to simplify the algebra before i integrate. Any step by step guidance would be appreciated

.

**Jester** · Nov 21st 2009, 09:29 AM

Originally Posted by purplerain

$\int _c xy^4ds$ C is the right half of the circle $x^2+y^2=16$

Here is what i did. Since C is the right half of the circle my bounds are from $-\pi/2 \leq t \leq \pi/2$ . I got x = 4cost and y = 4sint. I then take the arclength of that, and multiple that by the original function with the polar coordinates substituted into it which is $\int (4cost)(4sint)^4\sqrt{(-4sint^2)+(4cost^2)}dt$

Im kind of confused on how to simplify the algebra before i integrate. Any step by step guidance would be appreciated

.

So you have

$4^6\int_{-\pi/2}^{\pi/2} \cos t \sin^4 t \,dt$ then let $u = \sin t$ .

**purplerain** · Nov 21st 2009, 09:47 AM

Originally Posted by Danny

So you have

$4^6\int_{-\pi/2}^{\pi/2} \cos t \sin^4 t \,dt$ then let $u = \sin t$ .

I understand the integration, im just confused on how to simplfy the function before i integrate. The book has $4^5\int costsin^4t 4dt$

I have no idea how they got the 4dt, any help would be appreciated, my algebra is fuzzy.

**Jester** · Nov 21st 2009, 09:56 AM

Originally Posted by purplerain

I understand the integration, im just confused on how to simplfy the function before i integrate. The book has $4^5\int costsin^4t 4dt$

I have no idea how they got the 4dt, any help would be appreciated, my algebra is fuzzy.

$ x = 4 \cos t, y = 4\sin t \; \text{so}\; dx = -4 \sin t\, dt \; \text{and }\;dy = 4 \cos t\, dt $
so
$ ds = \sqrt{dx^2+dy^2}\,dt = \sqrt{(-4 \sin t)^2 + (4 \cos t)^2}\, dt$ $= \sqrt{4^2(\sin^2 t+ \cos^2t)}\,dt = \sqrt{4^2}\,dt = 4 dt $

**purplerain** · Nov 21st 2009, 10:51 AM

Originally Posted by Danny

$ x = 4 \cos t, y = 4\sin t \; \text{so}\; dx = -4 \sin t\, dt \; \text{and }\;dy = 4 \cos t\, dt $
so
$ ds = \sqrt{dx^2+dy^2}\,dt = \sqrt{(-4 \sin t)^2 + (4 \cos t)^2}\, dt$ $= \sqrt{4^2(\sin^2 t+ \cos^2t)}\,dt = \sqrt{4^2}\,dt = 4 dt $

Thank you!

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