- #1
Poop-Loops
- 732
- 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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