Question

The iterated integral \[\int_0^2\int_0^{4-y^2}\int_0^{2-x} dz\,dy\,dx\] represents the volume of which one of the following solid regions?

• A. The portion of the ellipsoid \(x^2/4+y^2/5+z^2/9\) above the plane z=4.
• B. The region bounded by the parabolic cylinder \(x=4-y^2\) and the plane \(z=2-x\).
• C. A paraboloid bounded by \(x^2+y^2=z\) and the plane z=4.
• D. The top half of a sphere \(x^2+y^2+z^2=25\) cut by the plane z=4.

Hint:

Read the z-limit as the upper bounding plane and the y-limit as tracing a parabolic cylinder in x.

Answer 1

The Correct Option is B

Step 1: Read off the three bounding surfaces from the limits. The outer variable x runs from 0 to 2. The middle bound ties y and x together through the relation \(x=4-y^2\), which is a parabolic cylinder in xyz-space (it does not involve z, so it extends as a cylindrical surface along the z-direction). The inner bound on z runs from the base \(z=0\) up to the plane \(z=2-x\).

Step 2: Interpret the solid geometrically. Over the base region carved out in the xy-plane by \(x=4-y^2\) (with \(0\le x\le 2\)), the solid is capped above by the slanted plane \(z=2-x\), which stays nonnegative exactly on this range since \(z=2-x\ge 0\) for \(x\le 2\). So the triple integral computes the volume of the solid lying between the parabolic cylinder \(x=4-y^2\) and the plane \(z=2-x\), sitting above \(z=0\).

Step 3: Eliminate the other options. There is no ellipsoid, circular paraboloid \(x^2+y^2=z\), or sphere anywhere in these limits; the bounding surfaces are strictly a parabolic cylinder in x and y, and a plane in x and z. Only one option names exactly these two surfaces. \[\boxed{\text{(B) The region bounded by the parabolic cylinder } x=4-y^2 \text{ and the plane } z=2-x}\]