Area enclosed by x=2cost, y=sint
By Anonymous on Thursday, February 8, 2001 - 02:06 pm :
A curve C is given by x=2cost, y=sint, 0 £ t < 2p, where t is a parameter. Find the area enclosed by C.

So x2=4cos2 t, and, y2=sin2 t

Because cos2 t+sin2 t=1,

Hence y2=1-(x2/4)

so y=[1-(x2/4)]1/2

when y=0, x=±2

what do I do next?

Thanks for any help.

By James Lingard (Jchl2) on Thursday, February 8, 2001 - 02:18 pm :
OK, the easiest way to do this equation is to rearrange the equation to

(x/2)2+y2=1

which is the equation for an ellipse, centered on the origin, with axes perpendicular to the x and y axes, and with height 4 and width 2. So the area enclosed by the ellipse is twice the area enclosed by the unit circle, so the area is 2p.

James

By Anonymous on Thursday, February 8, 2001 - 02:49 pm :

Thank you for that explanation James.

But how did you figure out that it was twice the area of the unit circle of just p?
By James Lingard (Jchl2) on Thursday, February 8, 2001 - 03:03 pm :
OK, what we've got is an ellipse, and ellipses are formed by taking circles and stretching them in one direction.

In our case, the ellipse is formed by taking the unit circle (with area p) and stretching it by a factor of 2 in the x-direction, so the resulting ellipse has area 2p.

If you know about matrices of transformations and their determinants, what we're doing is applying the transformation given by the matrix
æ
ç
è
1
0
0
2
 ö
÷
ø
which has determinant 2, so areas of things are multiplied by 2. If that means nothing to you then don't worry!

James

By Anonymous on Thursday, February 8, 2001 - 03:10 pm :

Thanks James for your explanation. I think I understand it better now.

Thanks again.