Volume of a body under a paraboloid inside a cylinder

This problem asks to find the volume of a body bounded by a paraboloid and a cylinder, using a double integral.

Problem

Find the volume of the body bounded by the surfaces:

z = x^{2} + y^{2}, z = 0, x^{2} + y^{2} = 1

Solution

Step 1

The projection of the body onto the Oxy plane is the unit disk D bounded by the cylinder x² + y² = 1. Above, the body is bounded by the paraboloid z = x² + y²; below, by the plane z = 0.

Why is the projection the unit disk?

The cylinder x² + y² = 1 is vertical (its axis is parallel to Oz), so the projection of the body onto Oxy coincides with its cross-section — the unit disk. The bounds in z are determined by the difference in heights between the paraboloid and the plane z = 0.

D : x^{2} + y^{2} \leq 1

Step 2

The volume of the body via a double integral of the difference in heights:

V = \iint_{D} (x^{2} + y^{2}) d x d y

Step 3

Change to polar coordinates: x = r cos φ, y = r sin φ, dx dy = r dr dφ:

Why are polar coordinates convenient?

The region D is a disk, and the integrand depends only on r² = x² + y². Polar coordinates turn the disk into the rectangle [0; 1] × [0; 2π] in (r, φ), and the integrand simplifies to r²·r = r³, which integrates elementarily.

x = r cos φ, y = r sin φ, x^{2} + y^{2} = r^{2}

d x d y = r d r d φ, 0 \leq φ \leq 2 π, 0 \leq r \leq 1

Step 4

Set the limits in the iterated integral:

V = \int_{0}^{2 π} d φ \int_{0}^{1} r^{2} \cdot r d r = \int_{0}^{2 π} d φ \int_{0}^{1} r^{3} d r

Step 5

Evaluate the inner integral over r:

\int_{0}^{1} r^{3} d r = \frac{r^{4}}{4} |_{0}^{1} = \frac{1}{4}

Step 6

Evaluate the outer integral over φ:

V = \int_{0}^{2 π} \frac{1}{4} d φ = \frac{1}{4} \cdot 2 π = \frac{π}{2}

Answer

V = π/2.

Frequently asked questions

How do I find the volume of a body using a double integral?

If the body is bounded above by a surface z = f(x, y) and below by a region D in the Oxy plane, its volume is V = ∬_D f(x, y) dx dy. If the body is also bounded below by another surface z = g(x, y), then V = ∬_D (f − g) dx dy.

When is it convenient to change to polar coordinates?

Polar coordinates are convenient when the integration region is a disk (or part of one), an annulus or a sector, and the integrand contains x² + y². Then dx dy = r dr dφ and x² + y² = r², simplifying the integral.

How do I choose the order of integration?

First integrate over the variable whose limits depend on the other. In polar coordinates, usually integrate over r first (from 0 to the boundary of the region), then over φ. In Cartesian coordinates — first over y (from the lower to the upper boundary), then over x.

What happens if the cylinder radius is doubled?

If the cylinder has radius R instead of 1, then V = ∫₀^{2π} dφ ∫₀^R r³ dr = 2π · R⁴/4 = πR⁴/2. For R = 1 we get π/2, for R = 2 — 8π. The volume grows as the fourth power of the radius.