Visualizing triple integrals

The solid region $\Omega$ is defined as the set of points $(x,y,z)$ such that $x^2+y^2 \le z \le 4$. It is enclosed below by the circular paraboloid $z=x^2+y^2$ and above by the plane $z=4$.

Conditions: The lower boundary is the paraboloid $z=x^2+y^2$.; The upper boundary is the plane $z=4$.; The region is bounded and closed.

The integral is evaluated by integrating with respect to $x$ and $y$ first over a horizontal disk cross-section, and then integrating with respect to $z$. Polar coordinates are used for the inner double integral, yielding an integrand of $r^3$.

Conditions: The solid region $\Omega$ is bounded below by the paraboloid $z=x^2+y^2$ and above by the plane $z=4$.; The integrand is $f(x,y,z) = x^2+y^2$.; The integration order is $dx\,dy\,dz$ (slicing method).

The final value of the triple integral is $\frac{32\pi}{3}$. This is obtained by evaluating the outer integral of $\frac{\pi}{2}z^2$ from $z=0$ to $z=4$.

Conditions: The solid region is bounded by $z=x^2+y^2$ and $z=4$.; The integrand is $x^2+y^2$.; The calculation follows the slicing method with polar coordinates for the inner integral.

Polar coordinates are used because both the boundary shape of the cross-section (a disk) and the integrand term $x^2+y^2$ exhibit rotational symmetry around the z-axis. This simplifies the algebra significantly, converting $x^2+y^2$ to $r^2$ and the area element to $r\,dr\,d\theta$.

Conditions: The cross-section $D(z)$ is a disk centered at the origin in the xy-plane.; The integrand contains the term $x^2+y^2$.

The radius $r$ of the cross-section disk is proportional to the square root of the height $z$. Specifically, $r = \sqrt{z}$.

Conditions: The solid is bounded below by the paraboloid $z=x^2+y^2$.; The cross-section is taken at a fixed height $z$.; The coordinate system is cylindrical/polar for the cross-section.

The horizontal cross-sections are disks. At any fixed height $z$, the intersection with the paraboloid $z=x^2+y^2$ creates a circular disk described by $x^2 + y^2 \le z$.

Conditions: The solid region is bounded below by $z=x^2+y^2$ and above by $z=4$.; The cross-sections are taken horizontally (perpendicular to the z-axis).