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
On
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
This is best computed using spherical coordinates... $$ \iiint_D (3+2xy) dV = \int_0^{2\pi}\int_0^{\pi/2} \int_0^2 r^2 \sin \varphi(3 + 2 r \sin \varphi \cos \theta \cdot r \sin \varphi \sin \theta)dr d \varphi d \theta = \cdots = 16 \pi $$