Evaluate the triple integrals over the indicated region.
$$\iiint_D (3+2xy)\,dV$$ over the solid hemispherical dome $D$ given by $x^2+y^2+z^2 \leq4$ and $z\geq0$.
The surface $3+2xy$ is above and surrounding the hemisphere. What volume would such integration calculate? How do I interpret this question? What is going on here? Sure I could calculate a volume bounded by some x and y values but in this case $0\leq Z\leq2$ but the surface $3+2xy$ barely lies in that $Z$ interval. I am confused
This is best computed using spherical coordinates... $$ \iiint_D (3+2xy) dV = \int_0^{2\pi}\int_0^{\pi/2} \int_0^2 r^2 \sin \varphi(3 + 2 r \sin \varphi \cos \theta \cdot r \sin \varphi \sin \theta)dr d \varphi d \theta = \cdots = 16 \pi $$
Firstly, the best way to interpret this problem would be my thinking about what the domain looks like in spherical coordinates. If we convert to we can see that: $$0\le r\le \theta$$ as this is simply the radius, now we want to rotated the full way around the "vertical" axis to get the whole xy-plane included, which means: $$0\le\theta\le 2\pi$$ and finally the second angle, now since we only want for $z\ge0$ we will only need: $$0\le\varphi\le\pi/2$$ Now just convert your formula, using the transformations from cartesian to polar coordinates and remember the $dV$ too and your all set
