Multiple Integrals in Polar and Cylindrical Coordinates

Rectangular limits fight back the moment a region is round. A quarter circle written in x and y hides a sqrt in its bounds, but written in polar it is just a box in r and theta. This lesson covers the change-of-coordinate machinery the MSTE paper actually tests: the r dr dtheta rule for double integrals in polar coordinates, area and volume over a polar region, triple integrals in cylindrical and spherical coordinates, the Jacobian that explains where every extra factor comes from, and centroids located by integration. Every rule is carried all the way to a number with units so you can check your own habits against it.

Double integrals in polar coordinates

A point in the plane can be named by its distance r from the origin and the angle theta it makes with the positive x-axis, with x = r cos theta and y = r sin theta. When you rewrite a double integral in these variables you cannot simply swap dx dy for dr dtheta, because a small polar cell is not a square. It is a little curved wedge whose two sides measure dr (radial) and r dtheta (along the arc), so its area is the product

dA = r dr dtheta.

That extra factor r is the whole game. It is small near the origin, where the wedges are pinched, and large far out, where the same angular span covers a wider arc. Forgetting it is the single most common polar mistake, and it always makes a circular area come out too small.

Worked example (area of a polar region): Find the area enclosed by the cardioid r = 1 + cos theta, with r in meters. The curve is traced once as theta runs from 0 to 2 pi, and the region reaches from the origin out to the boundary, so r runs from 0 to 1 + cos theta. The area is

A = integral from theta = 0 to 2 pi of integral from r = 0 to 1 + cos theta of r dr dtheta = integral from 0 to 2 pi of [r^2 / 2] from 0 to 1 + cos theta dtheta = (1/2) integral from 0 to 2 pi of (1 + cos theta)^2 dtheta.

Expanding, (1 + cos theta)^2 = 1 + 2 cos theta + cos^2 theta. Over a full turn the integral of 1 is 2 pi, the integral of 2 cos theta is 0, and the integral of cos^2 theta is pi (its average value 1/2 times the length 2 pi). So A = (1/2)(2 pi + 0 + pi) = 3 pi / 2 = 4.71 m^2. Dropping the factor of 1/2 that comes from the inner r-integration gives 3 pi = 9.42 m^2, the classic slip.

Area and volume over a polar region

Doing the inner r-integral in the area formula from 0 up to the boundary curve r(theta) gives the compact single-integral form that boards love:

A = (1/2) integral from alpha to beta of r^2 dtheta.

The same setup finds volume. If a surface z = f(x, y) sits above a polar region R, its volume is the double integral of the height times the base cell, and after substituting x = r cos theta and y = r sin theta the base cell is r dr dtheta:

V = double integral over R of f(x, y) dA = double integral over R of f(r cos theta, r sin theta) r dr dtheta.

Worked example (volume by a polar double integral): Find the volume of the solid dome bounded above by the upper hemisphere z = sqrt(9 - x^2 - y^2) and below by the disk of radius 3, all lengths in meters. In polar coordinates x^2 + y^2 = r^2, so the height is z = sqrt(9 - r^2), and the disk is 0 <= r <= 3, 0 <= theta <= 2 pi. Then

V = integral from 0 to 2 pi of integral from 0 to 3 of sqrt(9 - r^2) r dr dtheta.

For the inner integral substitute u = 9 - r^2, du = -2r dr, which turns it into [-(1/3)(9 - r^2)^(3/2)] from 0 to 3 = 0 - (-(1/3)(27)) = 9. So V = integral from 0 to 2 pi of 9 dtheta = 9(2 pi) = 18 pi = 56.55 m^3, exactly the (2/3) pi r^3 volume of a hemisphere of radius 3. Notice the r factor is what made the substitution clean; without it the inner integral has no elementary answer.

Triple integrals in cylindrical coordinates

Cylindrical coordinates are polar coordinates in the xy-plane with the ordinary z stacked on top: a point is (r, theta, z) with x = r cos theta, y = r sin theta, and z unchanged. Because only the plane part is bent, the volume element keeps the polar r factor and picks up dz untouched:

dV = r dz dr dtheta.

They are the right tool whenever the region has a circular cross-section that persists as you move along the axis: cylinders, cones, paraboloids, and pipes. As a quick sanity check, a solid cylinder of radius 2 and height 5 gives integral from 0 to 2 pi of integral from 0 to 2 of integral from 0 to 5 of r dz dr dtheta = (2 pi)(2)(5) = 20 pi = 62.83 cubic units, matching pi r^2 h = pi (2^2)(5) = 20 pi.

Worked example (volume by a triple integral in cylindrical coordinates): Find the volume of the solid bounded below by the plane z = 0 and above by the paraboloid z = 4 - x^2 - y^2, lengths in meters. In cylindrical form the cap is z = 4 - r^2, and it meets z = 0 where 4 - r^2 = 0, that is r = 2, so the shadow on the floor is the disk 0 <= r <= 2, 0 <= theta <= 2 pi. The height runs from z = 0 up to z = 4 - r^2. Then

V = integral from 0 to 2 pi of integral from 0 to 2 of integral from 0 to 4 - r^2 of r dz dr dtheta.

The inner z-integral of r dz from 0 to 4 - r^2 is r(4 - r^2) = 4r - r^3. The middle integral is integral from 0 to 2 of (4r - r^3) dr = [2r^2 - r^4 / 4] from 0 to 2 = (8 - 4) = 4. The outer integral is integral from 0 to 2 pi of 4 dtheta = 4(2 pi) = 8 pi = 25.13 m^3. Forgetting the r factor and integrating (4 - r^2) dz dr dtheta gives 32 pi / 3 = 33.51 m^3, a bit too large.

Triple integrals in spherical coordinates

Spherical coordinates name a point by its distance rho from the origin, the angle phi it makes down from the positive z-axis, and the same azimuth theta used before. The conversions are z = rho cos phi for the vertical part and, after projecting the horizontal reach rho sin phi onto the axes, x = rho sin phi cos theta and y = rho sin phi sin theta. A spherical cell has edges drho, rho dphi, and rho sin phi dtheta, so its volume is their product:

dV = rho^2 sin phi drho dphi dtheta.

To cover a whole solid ball, rho runs 0 to a, phi runs 0 to pi (north pole to south pole), and theta runs 0 to 2 pi. Swapping the phi and theta ranges is a frequent error that double counts one half of the ball and misses the other.

Worked example (volume of a sphere): Find the volume of a sphere of radius 3 m by direct integration. The three integrals separate because the limits are all constant:

V = integral from 0 to 2 pi of integral from 0 to pi of integral from 0 to 3 of rho^2 sin phi drho dphi dtheta = (integral from 0 to 3 of rho^2 drho)(integral from 0 to pi of sin phi dphi)(integral from 0 to 2 pi of dtheta).

The first factor is [rho^3 / 3] from 0 to 3 = 9. The second is [-cos phi] from 0 to pi = -(-1) + 1 = 2. The third is 2 pi. Multiplying, V = (9)(2)(2 pi) = 36 pi = 113.10 m^3, which is exactly (4/3) pi r^3 = (4/3) pi (3^3). Using pi rho^3 without the 4/3, or forgetting the sin phi factor, are the two errors to guard against.

The Jacobian of a transformation

Every extra factor above is one object: the Jacobian, the determinant that measures how much a coordinate change stretches or shrinks area or volume. For a plane change of variables x = x(u, v), y = y(u, v), the rule is

double integral over R of f dx dy = double integral over S of f |J| du dv, where J = (partial x / partial u)(partial y / partial v) - (partial x / partial v)(partial y / partial u).

Apply it to polar, x = r cos theta and y = r sin theta. The four partials are partial x / partial r = cos theta, partial x / partial theta = -r sin theta, partial y / partial r = sin theta, and partial y / partial theta = r cos theta, so

J = (cos theta)(r cos theta) - (-r sin theta)(sin theta) = r cos^2 theta + r sin^2 theta = r.

That is precisely the r in r dr dtheta. The same determinant, run in three dimensions, produces r for cylindrical and rho^2 sin phi for spherical. Memorize the three results and you never have to guess a volume element again.

Coordinate system	Element	Jacobian	Natural for
Polar (2D)	dA = r dr dtheta	r	disks, sectors, cardioids, anything circular
Cylindrical (3D)	dV = r dz dr dtheta	r	cylinders, cones, paraboloids, circular cross-sections along z
Spherical (3D)	dV = rho^2 sin phi drho dphi dtheta	rho^2 sin phi	balls, cones from the origin, radial fields

Worked example (change of variables): Evaluate the double integral of (x^2 + y^2) over the disk of radius 2 m. In rectangular coordinates the limits carry a sqrt; in polar they are a clean box. Since x^2 + y^2 = r^2 and dA = r dr dtheta,

double integral over R of (x^2 + y^2) dA = integral from 0 to 2 pi of integral from 0 to 2 of (r^2)(r) dr dtheta = integral from 0 to 2 pi of [r^4 / 4] from 0 to 2 dtheta = (16 / 4)(2 pi) = 4(2 pi) = 8 pi = 25.13 m^4.

Forgetting the r Jacobian collapses the inner integral to integral from 0 to 2 of r^2 dr = 8/3, giving 16 pi / 3 = 16.76 m^4, which is wrong by exactly the missing factor.

Centroids by integration

The centroid is the balance point of a uniform region or solid, found by dividing a first moment by the total area or volume. The moment about an axis is the integral of the perpendicular distance to that axis, and polar, cylindrical, or spherical coordinates simply make the integrals tractable when the shape is round.

Quantity for a region R or solid V	Integral form
Area A	double integral over R of dA
First moment about the x-axis, Mx	double integral over R of y dA
First moment about the y-axis, My	double integral over R of x dA
Centroid xbar	My / A
Centroid ybar	Mx / A
Solid centroid zbar	(1 / V) triple integral over V of z dV

Worked example (centroid of a semicircle): Find the centroid height of a uniform half disk of radius 6 cm with its straight edge on the x-axis. By symmetry xbar = 0, so only ybar is at stake. The region is 0 <= theta <= pi, 0 <= r <= 6, its area is A = (1/2) pi (6^2) = 18 pi, and y = r sin theta. The moment about the x-axis is

Mx = integral from 0 to pi of integral from 0 to 6 of (r sin theta) r dr dtheta = (integral from 0 to 6 of r^2 dr)(integral from 0 to pi of sin theta dtheta) = (72)(2) = 144.

So ybar = Mx / A = 144 / (18 pi) = 8 / pi = 2.55 cm, the standard 4a / (3 pi) result for a semicircle. Using the full-disk area pi a^2 by mistake would give 2a / (3 pi) = 1.27 cm, too low by half.

Worked example (centroid of a solid hemisphere): Find the centroid height of a uniform solid hemisphere of radius 4 cm resting with its flat face in the plane z = 0. Symmetry puts the centroid on the axis, and in spherical coordinates z = rho cos phi with the solid covering 0 <= rho <= 4, 0 <= phi <= pi/2, 0 <= theta <= 2 pi. The volume is V = (2/3) pi (4^3) = 128 pi / 3, and the moment about the z = 0 plane is

Mz = triple integral over V of z dV = (integral from 0 to 4 of rho^3 drho)(integral from 0 to pi/2 of cos phi sin phi dphi)(integral from 0 to 2 pi of dtheta) = (64)(1/2)(2 pi) = 64 pi.

Then zbar = Mz / V = 64 pi / (128 pi / 3) = 3(64) / 128 = 3/2 = 1.50 cm, the classic zbar = 3a / 8 for a solid hemisphere. Confusing it with the flat-lamina value 4a / (3 pi) = 1.70 cm is the usual trap.

Exam-day strategy

• Reach for polar the instant a boundary is a circle, a sector, or a cardioid: rewrite x^2 + y^2 as r^2, turn the boundary into r = something in theta, and set dA = r dr dtheta so the messy sqrt limits disappear.
• Never drop the Jacobian factor: it is r for both polar and cylindrical and rho^2 sin phi for spherical, and leaving it out always makes a round area or volume come out too small.
• For a full polar area use A = (1/2) integral of r^2 dtheta directly, and remember that the 1/2 is exactly the leftover from integrating r dr from 0 to the boundary.
• Match the coordinate system to the symmetry: cylindrical for circular cross-sections stacked along an axis (cylinders, cones, paraboloids), spherical for anything measured radially from a center (balls, spherical caps, cones from the origin).
• Set spherical limits carefully, with rho from 0 to a, phi from 0 to pi for a whole ball or 0 to pi/2 for the top half, and theta from 0 to 2 pi; keep the useful fact integral from 0 to pi of sin phi dphi = 2 ready.
• When the limits are all constant, split the triple integral into a product of three one-variable integrals and multiply, which is faster and far less error prone than nesting.
• For a centroid, compute the moment and the area or volume in whichever coordinates make the shape round, then divide; recall the two board staples ybar = 4a / (3 pi) for a semicircular area and zbar = 3a / 8 for a solid hemisphere.