# Area problem

• Mar 8th 2009, 04:27 PM
chengbin
Area problem
Determine the area in the 1st quadrant bounded by $\displaystyle x=acos^3t$, $\displaystyle y=asin^3t$, the x-axis and the y-axis (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.
• Mar 8th 2009, 08:29 PM
Mentia
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.
• Mar 9th 2009, 03:13 AM
chengbin
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.