91Ó°ÊÓ

A paraboloid is a three-dimensional shape that extends infinitely, resembling a stretched-out bowl or an upward-opening cup. In mathematics, a paraboloid can be described with specific properties, defined by an equation. In our exercise, the paraboloid is represented by the equation \(x^2 + y^2 = z\). This equation defines a paraboloid that opens upward, meaning it extends in the positive \(z\)-direction as both \(x\) and \(y\) increase.

To visualize this, imagine a 3D graph where planes parallel to the \(xy\)-plane cut horizontally through the paraboloid at different heights \(z\). Each slice creates a circular shape. The farther up the \(z\)-axis you go, the larger the circle becomes, as it encompasses more of the paraboloid's surface. Understanding this key property helps in solving problems where planes intersect with such surfaces.

When a plane intersects a paraboloid, it forms a cross-section that can often be a circle. The intersection depends on where the plane cuts the paraboloid. In our scenario, two planes, \(z = 2\) and \(z = 6\), intersect the paraboloid \(x^2 + y^2 = z\).

To find where exactly these planes slice through, we substitute the values of \(z\) back into the paraboloid equation. For \(z=2\):
- Substitute to get \(x^2 + y^2 = 2\). This represents a circle with radius \(\sqrt{2}\), calculated by taking the square root of both sides.
For \(z=6\):
- Substitute to get \(x^2 + y^2 = 6\). This results in a circle with radius \(\sqrt{6}\).
These radii show how these circles differ in size, with the circle at \(z=6\) being larger due to the higher intersection point on the paraboloid.

The area between two circular intersections of planes with a paraboloid is known as a circular band. This band lies between the two circles sliced by planes at different heights on the paraboloid surface.

To calculate this band area, we use the formula for the area of a circle, \(\pi r^2\), and compute individual areas of the circles formed at \(z=2\) and \(z=6\).

• The area of the larger circle (at \(z=6\), radius \(\sqrt{6}\)) is \(6\pi\) since \((\sqrt{6})^2 = 6\).
• The area of the smaller circle (at \(z=2\), radius \(\sqrt{2}\)) is \(2\pi\) as \((\sqrt{2})^2 = 2\).

By subtracting the area of the smaller circle from the larger one \((6\pi - 2\pi)\), we find that the area of the circular band is \(4\pi\).

This difference in areas truly represents the region of the paraboloid surface that lies between the two planes. Understanding this concept is vital for solving more complex problems involving cylindrical or spherical intersections with other surfaces.