To find the area of a given curve.

84 Views Asked by At

The curve is given $$x =(1-t^{3})/(1+t^{2})$$ and $$y= 2t/(1+t^{2})$$

I know the method for finding the area, but I'm having problem with the tracing of curves. In exams, I won't really have time to trace the curve by finding values of $x$ and $y$ using values of $t$.

I just want to have a rough idea about the curve, so that I can find the values of $t$ for which the curve makes a complete loop.

How to do it?

1

There are 1 best solutions below

2
On

Hint:

The curve $y(x)$ does not form a loop itself. It is the area enclosed by $y(x)$, the y-axis and the x-axis. Then, it is easy to verify that the curve intersects the x-axis at (1,0) and y-axis at (0,1). As a result, the area

$$I = \int_0^1 y(x)dx$$

can be carried out with the variable substitutions $x=(1-t^3)/(1+t^2)$ and $y=2t/(1+t^2)$. Explicitly,

$$I=\int_0^1 \frac{2t}{1+t^2}\left[1-\frac{1}{1+t^2}+\frac{2t}{(1+t^2)^2}\right]dt$$

which can then be integrated piecewise.

enter image description here