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Change the Cartesian integral to an equivalent polar integral, and then evaluate.
∫₋₁₁¹¹ ∫₀√⁽¹²¹⁻ˣ²⁾ dy dx

\int\limits^{11}_{-11} \int\limits^{\sqrt{121-x^{2}} } _{0} dy dx

Sagot :

It looks like the starting integral is

[tex]\displaystyle \int_{-11}^{11} \int_0^{\sqrt{121-x^2}} dy\,dx[/tex]

The key is recognizing the domain of integration as the semicircle centered at the origin with radius 11. In polar coordinates, this integral transforms to

[tex]\displaystyle \boxed{\int_0^\pi \int_0^{11} r \, dr \, d\theta}[/tex]