(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I dont really have a problem with integration here, I just need to learn how to decide what direction the integration should be in

find the area enclosed by the curve x = t^{2}- 2t y = t^{1/2}around the y axis

2. Relevant equations

A = ∫ xdy

3. The attempt at a solution

so when i plug values into the parametric equations i find that this graph comes out of the origin at y = 0 then crosses the axis again at 2^{1/2}

I feel like my bound of integration should be from 0 to 2^{1/2}, but I get a negative area, so I probably should reverse the bounds, but I cant rationalize the reversal.

A = from 0 to 2^{1/2}∫ (t^{2}- 2t)t^{1/2}dt

= from 0 to 2^{1/2}( 4/5 *t^{5/2 }- 4/3 * t^{3/2})

here i get 1.86 - 2.2, which is negative

**Physics Forums - The Fusion of Science and Community**

# Find the area enclosed by the curve (parametric equation)

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Find the area enclosed by the curve (parametric equation)

Loading...

**Physics Forums - The Fusion of Science and Community**