Çözüldü Circle Equation - Integral

The changed and slightly challenged version of an extremely easy question:

By integrating, compute the area of the region bounded by the circle that is tangent to both axes with a radius of 6 units, and the coordinate axes in the third quadrant.

From the equation in the question [ x - (-6) ]^2 + [ y - (-6) ]^2 = 6^2 ⇒ y = f(x) = ( 36 - (x + 6)^2 )^0,5 - 6
Area = (lower bound x = -6, upper bound x = 0), ∫ ( 0 - ( ( 36 - (x + 6)^2 )^0,5 - 6 ) ) dx
Area = (lower bound x = -6, upper bound x = 0), ∫ ( 6 - ( 36 - (x + 6)^2 )^0,5 ) dx

https://i72.servimg.com/u/f72/19/97/10/39/area_c10.png
https://www.wolframalpha.com/input?i=integrate (6-sqrt(36-(x+6)^2)) dx from x=-6..0
Note: Homework for the interested students to calculate and find the correct result using variable conversion x + 6 = 6sin(θ) or
x + 6 = 6cos(θ).

Graph:

https://i72.servimg.com/u/f72/19/97/10/39/area_c11.png

Geometry Solution: 6·6 - (π·6^2 / 4) = 36 - 9π Unit^2.

Original Question (Turkish):
https://ogmmateryal.eba.gov.tr/soru...,6025,6027,6028,6034&p=6&t=css&ks=0&os=0&zs=0
(Son soru, "Testi Tamamla" butonuna tıklanınca yanıtlar görünüyor.)