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## Homework Statement

Where the region is:

D = {(x,y)| 0[tex]\leq[/tex]x[tex]\leq[/tex]2;0[tex]\leq[/tex]y[tex]\leq[/tex][tex]\sqrt{}2x-x^2[/tex]}

Double integral over region D with f(x,y) = [tex]\sqrt{}x^2+y^2[/tex] and respect to dA

## Homework Equations

Trig. Identities:

x = rcos(theta)

y = rsin(theta)

x^2+y^2 = r^2

## The Attempt at a Solution

First, I graphed D in xy-plane and got a semi-circle with its center translated to the right by 1 units and is bounded by

x=0 and x=2

I converted to polar coordinates and got these:

f(x,y) = r

dA = r(dr)d(theta)

D = {(r,theta)| 0[tex]\leq[/tex]r[tex]\leq[/tex]2;x/0[tex]\leq[/tex]theta[tex]\leq[/tex]pi}

I'm not sure if everything is accounted from (x,y) to (r,theta) and the cone that is made from f(x,y).

Thank you for taking your time to read this.