Define
x = r Cos[t];
y = r Sin[t];
Then
x^2 + y^2 // Simplify
r^2
So your r is equal to 6.
Your constraints are fulfilled if 3Pi/2 <= t <= 2 Pi. Look for example at
Plot[{x /. r -> 1, y /. r -> 1.5}, {t, 0, 2 Pi}]
The Volumeelement in Polar Coordinates (in R2 ) is dV = r dr dt, so your integral is written as
In[68]:= 1 / Pi Integrate[ r^2 r , {r, 0, 6}, {t, 3 Pi/2, 2 Pi}]
Out[68]= 162
Obviously your integral "lives" on a disk with radius r. Then you can equally write y as function of x and your integral is as well
In[77]:= 1/Pi Integrate[ x^2 + y^2, {x, 0, 6}, {y, 0, Sqrt[36 - x^2]}]
Out[77]= 162
