# Area of parametric curve

• Mar 22nd 2009, 11:11 AM
triggy
Area of parametric curve
Hi, I'm having trouble finding the area of this parametric curve.

The area is enclosed by the curve x = t^2-2t, y = t^0.5 and the y-axis.

Thanks.
• Mar 22nd 2009, 11:52 AM
skeeter
Quote:

Originally Posted by triggy
Hi, I'm having trouble finding the area of this parametric curve.

The area is enclosed by the curve x = t^2-2t, y = t^0.5 and the y-axis.

Thanks.

sketch a graph ... note that the area will be an integral w/r to y.

$\displaystyle A = \int_c^d x(t) \cdot y'(t) dt$

$\displaystyle y = \sqrt{t}$

$\displaystyle dy = \frac{1}{2\sqrt{t}} \, dt$

$\displaystyle x = 0$ at $\displaystyle t = 0$ and $\displaystyle t = 2$ ...

$\displaystyle A = \int_0^2 (t^2 - 2t) \cdot \frac{1}{2\sqrt{t}} \, dt$

can you finish from here?
• Mar 22nd 2009, 12:21 PM
triggy
Oh alright, I got up to that last line.

I integrated and got an answer of (1/5)*[2^(5/4)]/5 - (2/3)*[2^(3/4)].

It doesn't appear to be the correct answer. I was wondering about the bounds. Is the bounds supposed to be in terms of t: the integral of 0 to 2?
• Mar 22nd 2009, 12:28 PM
skeeter
Quote:

Originally Posted by triggy
Oh alright, I got up to that last line.

I integrated and got an answer of (1/5)*[2^(5/4)]/5 - (2/3)*[2^(3/4)].

It doesn't appear to be the correct answer. I was wondering about the bounds. Is the bounds supposed to be in terms of t: the integral of 0 to 2?

you're right, it should be 0 to 2 ... my mistake.