
Area problem
Determine the area in the 1st quadrant bounded by $\displaystyle x=acos^3t$, $\displaystyle y=asin^3t$, the xaxis and the yaxis (a>0)
In my solution book it says:
As x varies from 0 to a, t varies from $\displaystyle \frac {\pi} {2}$ to 0
How do you know that?
Skipping all the way down to this step
$\displaystyle 3a^2\int_\frac{\pi} {2} ^2 sin^4tdt  3a^2\int_\frac{\pi} {2} ^2 sin^6tdt$
= $\displaystyle [3a^2*(\frac{1} {2}*\frac{3} {4}*\frac{\pi} {2}]  [3a^2*(\frac{1} {2}*\frac{3} {4}*\frac{5} {6}*\frac{\pi} {2}]$
How do you get to that step?
Thanks a lot.

So what they did was transform to a coordinate system that is easier to integrate.
You can just plug in values to see what happens to x as t varies, it shows that as x varies 0>a, t varies 0>Pi/2.
We know $\displaystyle x=aCos(t)^{3}$, so
$\displaystyle \left  \frac {dx} {dt} \right  =3aCos(t)^{2}Sin(t)$, this is our Jacobian.
Then we have:
Area = $\displaystyle \int_{ 0}^{ \frac{ \pi }{2 } } \int_{0 }^{aSin(t)^{3} } 3aCos(t)^{2}Sin(t)dydt = \int_{0}^{ \frac{\pi}{2}} 3a^{2} \cos ^2(t) \sin ^4(t) dt$
But we have cos(t)^2 = 1  sin(t)^2 which gives:
$\displaystyle \int_{0}^{ \frac{\pi}{2}} 3a^{2} \sin ^4(t)dt  \int_{0}^{ \frac{\pi}{2}} 3a^{2} \sin ^6(t)dt$
I'm not 100% sure why their limits are Pi/2 to 2 instead of my 0 to Pi/2 but it doesn't seem to change the answer.

I've never learned double integrals before.
I know I'm supposed to plug in values from 0 to a, but what values? What is the limit of a? (a<1?, a<2? a<3?)
I get this step
$\displaystyle
3a^2\int_\frac{\pi} {2} ^2 sin^4tdt  3a^2\int_\frac{\pi} {2} ^2 sin^6tdt
$
But I don't get the step after that.