Find the area enclosed by the curve (parametric equation)

[vande060]

Homework Statement

I don't 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 = t² - 2t y = t^1/2 around the y axis

Homework Equations

A = ∫ xdy

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 can't rationalize the reversal.

A = from 0 to 2^1/2 ∫ (t² - 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

 

[SammyS]

Have you looked at the graph?

 

[vande060]

sure it looks like this approx

starting at t=0 there the curve is at the origin, and at point t = 2 the graph it at the point 0,2^1/2

so the bounds i used seemed right

 

[eczeno]

the area should be negative; it is 'below' the y-axis.

there is one mistake that i can see. if y = t^(1/2), then dy = (1/2)*t^(-1/2) dt. you seem to have integrated xy instead of xdy.

 

[vande060]

the area should be negative; it is 'below' the y-axis.

there is one mistake that i can see. if y = t^(1/2), then
dy = (1/2)*t^(-1/2) dt. you seem to have integrated xy instead of xdy.

that clears everything up thank you,

 

[eczeno]

cheers