Assume that the integral ${\displaystyle \underset{0}{\overset{6}{}}}{\displaystyle \underset{-\sqrt[\phantom{2}]{36-y^2}}{\overset{\sqrt[\phantom{2}]{36-y^2}}{}}}f(x,y)dxdy$ exists.

Which is the equivalent double integral with the order of integration changed?

${\displaystyle \underset{-\sqrt[\phantom{2}]{36-y^2}}{\overset{\sqrt[\phantom{2}]{36-y^2}}{}}}{\displaystyle \underset{0}{\overset{6}{}}}f(x,y)dydx$

${\displaystyle \underset{0}{\overset{6}{}}}{\displaystyle \underset{-\sqrt[\phantom{2}]{36-x^2}}{\overset{\sqrt[\phantom{2}]{36-x^2}}{}}}f(x,y)dydx$

${\displaystyle \underset{-6}{\overset{6}{}}}{\displaystyle \underset{0}{\overset{\sqrt[\phantom{2}]{36-x^2}}{}}}f(x,y)dydx$

${\displaystyle \underset{0}{\overset{6}{}}}{\displaystyle \underset{\sqrt[\phantom{2}]{36-x^2}}{\overset{36}{}}}f(x,y)dydx$

${\displaystyle \underset{-6}{\overset{6}{}}}{\displaystyle \underset{-\sqrt[\phantom{2}]{36-x^2}}{\overset{\sqrt[\phantom{2}]{36-x^2}}{}}}f(x,y)dydx$