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**[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!

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