Question is: Using polar coordinates, evaluate the integral $\iint\sin(x^2+y^2)dA$ where R is the region $1\le x^2+y^2\le81$
My work for the inside integral, using bounds 1 to 9, was a constant:
$$\int_1^9\sin(r^2)rdr=-\frac{1}{2}\left(\cos81-\cos1\right)$$
Does this mean I can conclude my answer is this constant? In other words, can I bring it out of the integral with respect to theta from 0 to 2$\pi$? I am guessing not because my answer was not correct. If not, could someone explain why?
You ignored the angular part of the integral: $$\int\int_R\sin(x^2+y^2)\,dx\,dy=\int_0^{2\pi}\int_1^9r\sin(r^2)\,dr\,d\theta\\=\left(\int_0^{2\pi}\,d\theta\right)\left(\int_1^9r\sin(r^2)\,dr\right)=2\pi\times(-0.118\dots)=-0.74\dots$$