Transform to polar coordinates

[dimi212121]

Could someone please convert this double integral to polar coordinates?
0<x<1, x*2<y<1 Int.Int f(x,y)dxdy

[Defennder]

What is x*2?

[dimi212121]

Sorry x^2

[Zizy]

If you draw your area, you ll see that it looks like 1/4 of a circle. Use something like that:
x=1+rcos(fi), y=1+rsin(fi). R will be going from 0 to 1, fi from -pi/2 to 0 then. dxdy gets changed to rdrdfi.
This is not the only solution of course.

[tiny-tim]

Could someone please convert this double integral to polar coordinates?
0<x<1, x*2<y<1 Int.Int f(x,y)dxdy

(have an int: ∫ and a squared: ² and a theta: θ and a pi: π :smile:)

The area is a triangle for π/4 < θ < π/2.

For 0 < θ < π/4, you simply need to write the boundary as an equation in r and θ.

Hint: for 0 < θ < π/4, each line of constant θ has tanθ = y/x, and it hits the boundary at y = x².

So what is the value of x at the boundary … and what value of r does that correspond to? :smile: