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Area under astroid

[solved]Area under astroid

1. Homework Statement
So this is another one of those parametric questions that I don't know what I did wrong.

Determine the area of the region enclosed by the astroid

[tex]x(t) = 8cos^{3}t[/tex]
[tex]y(t) = 8sin^{3}t[/tex]



2. Homework Equations

[tex] Area = \int y dx [/tex]

3. The Attempt at a Solution


Before I start I must admit that I don't have a good knowledge of this topic of area under parametric curves. My calc textbook had half a page on it and just one example so I am rather clueless.

[tex] Area = 2\int^{8}_{0}ydx [/tex] (Instead of doing -8 to 8, I did 0 to 8 since it is symmetric)

Ok, so y is already given so I just need to find dx.

[tex]dx = -24cos^{2}tsint dt[/tex]


[tex] Area = 2\int^{8}_{0}8sin^{3}t( -24cos^{2}tsint dt)[/tex]
[tex] Area = 2\int^{8}_{0}-192sin^{4}tcos^{2}tdt[/tex]
[tex] Area = -384\int^{8}_{0}sin^{4}tcos^{2}tdt[/tex]
[tex] Area = -384\int^{8}_{0}(sin^{2}t)^{2}cos^{2}tdt[/tex]
[tex] Area = -384\int^{8}_{0}( \frac{1-cos(2t)}{2})^{2}(\frac{1+cos(2t)}{2})dt[/tex]

So anyhow.. there were alot of integration steps and I arrived at

[tex]Area = -24\left[\frac{-1}{3}sin^{3}(2t)+t-\frac{1}{2}sin4t\right]^{8}_{0} [/tex]


When I evaluate that, it gave me -185.5737904 when the answer is suppose to be 75.3982236

Thanks for any help!
 
Last edited:

Answers and Replies

rock.freak667
Homework Helper
6,230
31
Your limits are x=8 and x=0 which is true when you integrate with respect to x. You are integrating with respect to t.
 
dynamicsolo
Homework Helper
1,648
4
When you make the change in the limits of integration to go with your substituted variable, you'll also find that the order of integration reverses. Reversing the order again to get the integration running in the right direction will take care of the negative sign you're getting for your area.
 
Hey guys thanks for the help. I figured it a bit earlier that I was doing it entirely wrong heh. Thanks though.
 

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