- #1

Poop-Loops

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- 1

So a previous problem said to show that the area of some simple closed curve C was:

[tex]A= \frac{1}{2} \oint_{C} (xdy - ydx)[/tex]

Simple enough. My problem says to find the area of the curve

[tex]x^{2/3} + y^{2/3} = 4[/tex]

using that formula.

So off to polar coordinate land I go. r = 4, x = cos(t), dx = -sin(t), y = sin(t), dy = cos(t), right?

So after plugging everything in and taking out the r, I get:

[tex]A = 2 \oint_{0}^{2 \pi} (cos(t)^{5/3} + sin(t)^{5/3})dt[/tex]

Correct?

How do I integrate that? They aren't squared. And I can't find any trig formula to simplify that or any integration formula. Do I have to venture into infinite series?

[tex]A= \frac{1}{2} \oint_{C} (xdy - ydx)[/tex]

Simple enough. My problem says to find the area of the curve

[tex]x^{2/3} + y^{2/3} = 4[/tex]

using that formula.

So off to polar coordinate land I go. r = 4, x = cos(t), dx = -sin(t), y = sin(t), dy = cos(t), right?

So after plugging everything in and taking out the r, I get:

[tex]A = 2 \oint_{0}^{2 \pi} (cos(t)^{5/3} + sin(t)^{5/3})dt[/tex]

Correct?

How do I integrate that? They aren't squared. And I can't find any trig formula to simplify that or any integration formula. Do I have to venture into infinite series?

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