Area problem

**chengbin** · March 8th 2009, 04:27 PM

Determine the area in the 1st quadrant bounded by $x=acos^3t$ , $y=asin^3t$ , 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 $\frac {\pi} {2}$ to 0

How do you know that?

Skipping all the way down to this step

$3a^2\int_\frac{\pi} {2} ^2 sin^4tdt - 3a^2\int_\frac{\pi} {2} ^2 sin^6tdt$

= $[3a^2*(\frac{1} {2}*\frac{3} {4}*\frac{\pi} {2}] - [3a^2*(\frac{1} {2}*\frac{3} {4}*\frac{5} {6}*\frac{\pi} {2}]$

How do you get to that step?

Thanks a lot.

**Mentia** · March 8th 2009, 08:29 PM

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 $x=aCos(t)^{3}$ , so

$\left | \frac {dx} {dt} \right | =3aCos(t)^{2}Sin(t)$ , this is our Jacobian.

Then we have:

Area = $\int_{ 0}^{ \frac{ \pi }{2 } } \int_{0 }^{aSin(t)^{3} } 3aCos(t)^{2}Sin(t)dydt = \int_{0}^{ \frac{\pi}{2}} 3a^{2} \cos ^2(t) \sin ^4(t) dt$

But we have cos(t)^2 = 1 - sin(t)^2 which gives:

$\int_{0}^{ \frac{\pi}{2}} 3a^{2} \sin ^4(t)dt - \int_{0}^{ \frac{\pi}{2}} 3a^{2} \sin ^6(t)dt$

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.

**chengbin** · March 9th 2009, 03:13 AM

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

$<br /> 3a^2\int_\frac{\pi} {2} ^2 sin^4tdt - 3a^2\int_\frac{\pi} {2} ^2 sin^6tdt <br />$

But I don't get the step after that.

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