# Homework Help: Caclulus with parametric curves

1. May 24, 2010

### nameVoid

Elipse [0,2pi]
x=acosT
y=bsinT\
by symmetry
4I(y,x)
chossing the limits of integration my text is showing
4I(y,x,0,a)
T=0=>x=a
T=pi/2=>x=0
so wouldnt this be 4I(y,x,a,0)
proceeding inthe books case gives pi(ab) while in the later case gives -pi(ab)

Also
x= (t-1)/t y=(t+1)/t bounded by y=5/2
5/2=(t+1)/t => t=2/3
my text is showing t=1/2 or 2

Last edited: May 24, 2010
2. May 24, 2010

### HallsofIvy

It would have helped if had explained what you are talking about! I think you are trying to find the area of an ellipse and are saying that "by symmetry, the area of the ellipse is 4 times the area the part of the ellipse in the first quadrant".

Limits of integration for what integral? Since you are using parametic equations with a single parameter, you appear to be integrating around the circumference of the ellipse. But that won't give "area".

I guessed, above, that "I(y,x)" was the area of the part of the ellipse in the first quadrant but now I have no idea what "I(y,x,a,0)" could mean.

The are of an ellipse with semi-axes a and b is $\pi ab$. Obviously, an area cannot be negative.

Please tell us what the problem really is!

I have no idea what you mean by "showing t= 1/2 or 2". Nor do I understand what is meant by "bounded by y= 5/2". Those parametric equations describe a curve. A curve has to be bounded at both ends.