Seeing that this is my first time posting, I hope I'm following the rules correctly. Anyways, the question I'm stuck on is:
Evaluate:
$$ \int_{-3}^3\int_{-2}^2\int_{-\sqrt{9-y^2}}^\sqrt{9+y^2} 13^{x^2+y^2}x + 2zx^{y^{x}}-3 \space dx \space dz \space dy $$
So far, I've thought of dividing up the integral into three (or rather, two serious and one elementary part, as follows:
$$ \int_{-3}^3\int_{-2}^2\int_{-\sqrt{9-y^2}}^\sqrt{9+y^2} 13^{x^2+y^2}x \space dx \space dz \space dy + \int_{-3}^3\int_{-2}^2\int_{-\sqrt{9-y^2}}^\sqrt{9+y^2} 2zx^{y^{x}} \space dx \space dz \space dy - \int_{-3}^3\int_{-2}^2\int_{-\sqrt{9-y^2}}^\sqrt{9+y^2} 3 \space dx \space dz \space dy $$
And to solve the first part, I changed the coordinate system as follows:
$$ \int_{0}^{2\pi}\int_{-2}^2\int_{0}^3 r\cos(\theta) 13^{r^2} r \space dr \space dz \space d\theta \Rightarrow 0 \space [\because of \cos(\theta)] $$
And, that's it. I can't seem to think of a way to evaluate the other part of the integral. I tried to think of any advantage I could get by changing the order of integration, i.e. dx first or dy or dz and so on and so forth, but it doesn't seem to be of any apparent convenience. Any suggestions?
The second part of your integral is an odd function with respect to $z$. Also, for the third part, notice that integrating a constant over a region just gives the volume of the region multiplied by the constant.
More explicitly we have:
$$\iiint2zx^{y^x}dV=\int_{-2}^2zdz\cdot\iint x^{y^x}dA=0\cdot\iint x^{y^x}dA=0$$