What is the area under the astroid curve?

[motornoob101]

[solved]Area under astroid

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

x(t) = 8cos^{3}t
y(t) = 8sin^{3}t

Homework Equations

Area = \int y dx

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.

Area = 2\int^{8}_{0}ydx (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.

dx = -24cos^{2}tsint dt


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

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

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


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

Thanks for any help!

 

[rock.freak667]

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]

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.

 

[motornoob101]

Hey guys thanks for the help. I figured it a bit earlier that I was doing it entirely wrong heh. Thanks though.