Use of integration to find area

[Calpalned]

Homework Statement

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

Homework Equations

Area of a parametric curve = ∫g(t) f'(t) dt, where g(t) = y and f(t) = x

The Attempt at a Solution

I believe that the limits of integration by be found by setting x and y equal to each other and solving for y. Do I still use the given equation, or do I have to modify it as this question asks about the y-axis, not the x-axis.

Thanks

 

[Dick]

Homework Statement

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

Homework Equations

Area of a parametric curve = ∫g(t) f'(t) dt, where g(t) = y and f(t) = x

The Attempt at a Solution

I believe that the limits of integration by be found by setting x and y equal to each other and solving for y. Do I still use the given equation, or do I have to modify it as this question asks about the y-axis, not the x-axis.

Thanks

The limits of integration are the values of t where your curve intersects the y-axis. The y-axis is where the value of x equals 0. Solve that.

 

[Calpalned]

I solved t^2 - 2t = x = 0 and I got t = 2 and t = 0, so those are my limits
Now, do I still use ∫g(t) f'(t) dt or do I have to modify the equation to ∫f(t) g'(t) dt ?

 

[Dick]

I solved t^2 - 2t = x = 0 and I got t = 2 and t = 0, so those are my limits
Now, do I still use ∫g(t) f'(t) dt or do I have to modify the equation to ∫f(t) g'(t) dt ?

You can actually use either one. One will give you the negative of the other. You can see this by looking at integration by parts.