I have this integral: $$\int\int_D(x^3y^2+\ln(x^2+x+1)\sin(y^3))dA$$ Where $D: x^2+y^2 \le 1$ I was wondering what the best way to solve this using symmetry rather than using calculations? Any hints or tips would be appreciated!
The function $$f(x,y) =(x^3y^2) +(\ln(x^2+x+1)\sin(y^3)) $$ The first part namely $f_1(x,y)=(x^3y^2)$ satisfies $$f_1(x,y)=-f_1(-x,y)$$
The second part namely $$f_2(x,y) = (\ln(x^2+x+1)\sin(y^3))$$ satisfies $$f_2(x,y)=-f_1(x,-y)$$
Since the region is symmetric with respect to both $x-axis$ and $y-axis$ the positive and negative contributions cancel each other and you get $$\int\int_D(x^3y^2+\ln(x^2+x+1)\sin(y^3))dA=0$$
answered Oct 8, 2019 at 21:24 Mohammad Riazi-Kermani Mohammad Riazi-Kermani 68.9k 4 4 gold badges 42 42 silver badges 89 89 bronze badges$\begingroup$ Your first statement is not correct, right? $f(x,-y)\neq f(x,y)$ in general (note the first term doesn't change sign with $y$ but the second term does, since $y^2$ is even but $\sin y^3$ is odd). Is that really what you meant? $\endgroup$