$\int\int\int_{V}(x-y)dV$ where $V$ is volume enclosed by : $ S=\left\{(x,y,z):(x^{2}+y^{2})^{2}+z^{4}=16;z\geq0\right\}$
What I did: $\int\int\int_{V}(x-y)dV=\int\int_{A}\left[\int_{0}^{\left(16-(x^{2}+y^{2})^{2}\right)^{1/4}}(x-y)dz\right]dA=\int\int_{A}(x-y)\left(16-(x^{2}+y^{2})^{2}\right)^{1/4}dA$ where $A=\{(x,y):x^{2}+y^{2}\leq4$ I tryed changing to polars next, but didn't helped much... I don't think it's relevant, but $(x-y)$ is a $div (f)$ that I got earlyer.