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Step 18 · Multivariable calculus topics

Switch to coordinates that fit the region, such as polar, cylindrical or spherical, and use the Jacobian to adjust the area or volume element.

Polar coordinates in integrals

For circles, discs and wedges, use x = r cos θ and y = r sin θ. A small polar patch is nearly a rectangle with sides dr and r dθ, so dA = r dr dθ.

∬Rf(x,y) dA=∫αβ∫abf(rcos⁡θ,rsin⁡θ) r dr dθ\iint_R f(x, y)\,dA = \int_{\alpha}^{\beta}\int_{a}^{b} f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta

Example: Area of a disc

Find the area of the disc of radius 3 with a polar integral.

  1. ∫02π∫03r dr dθ=2π⋅92\int_{0}^{2\pi}\int_{0}^{3} r\,dr\,d\theta = 2\pi \cdot \frac{9}{2}.

Answer9π

Example: A bell-shaped integral

Find ∬De−(x2+y2) dA\iint_D e^{-(x^{2} + y^{2})}\,dA over the disc of radius 1.

  1. In polar form it is ∫02π∫01e−r2 r dr dθ\int_{0}^{2\pi}\int_{0}^{1} e^{-r^{2}}\,r\,dr\,d\theta.
  2. The inner integral is 12(1−e−1)\frac{1}{2}\left(1 - e^{-1}\right).

Answerπ(1−e−1)\pi\left(1 - e^{-1}\right), about 1.99

Cylindrical and spherical coordinates

Cylindrical coordinates add z to polar coordinates: good for cylinders and cones. Spherical coordinates use the distance ρ from the origin, the angle φ down from the z-axis and the angle θ around it: good for balls.

SystemCoordinatesVolume element
Cylindricalr, θ, zr dz dr dθr\,dz\,dr\,d\theta
Sphericalρ, φ, θρ2sin⁡φ dρ dφ dθ\rho^{2}\sin\varphi\,d\rho\,d\varphi\,d\theta

Example: A cylinder

Find the volume of a cylinder of radius 2 and height 5 with cylindrical coordinates.

  1. ∫02π∫02∫05r dz dr dθ\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{5} r\,dz\,dr\,d\theta.
  2. The three factors are 5, 2 and 2π.

Answer20π

Example: A ball

Find the volume of a ball of radius R with spherical coordinates.

  1. ∫02π∫0π∫0Rρ2sin⁡φ dρ dφ dθ\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R} \rho^{2}\sin\varphi\,d\rho\,d\varphi\,d\theta.
  2. The three factors are R33\frac{R^{3}}{3}, 2 and 2π.

Answer43πR3\frac{4}{3}\pi R^{3}

The Jacobian

A general change of variables x = x(u, v), y = y(u, v) stretches small squares into small parallelograms. The Jacobian determinant gives the stretch factor, so dA = |J| du dv.

J=∂(x,y)∂(u,v)=∣xuxvyuyv∣J = \frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} x_u & x_v \\ y_u & y_v \end{vmatrix}

Example: Polar, as a check

Find the Jacobian of x = r cos θ, y = r sin θ.

  1. J = cos θ · r cos θ − (−r sin θ) · sin θ.
  2. That is r(cos²θ + sin²θ).

Answerr: the extra r in polar integrals

Example: A slanted region

Use u = x + y and v = x − y to find the area of the square with corners (0, 0), (1, 1), (2, 0) and (1, −1).

  1. In u and v the region is the square 0 ≤ u ≤ 2, 0 ≤ v ≤ 2.
  2. x = (u + v) ÷ 2 and y = (u − v) ÷ 2, so |J| is one half.
  3. The area is one half of 4.

Answer2

Try it yourself

Work each one on paper first, then open the answer to check.

  1. What is dA in polar coordinates?

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    r dr dθ

  2. Find ∫02π∫02r dr dθ\int_{0}^{2\pi}\int_{0}^{2} r\,dr\,d\theta.

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    4π

  3. Find the area inside r = 1 for 0 ≤ θ ≤ π/2 with a polar integral.

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    π4\frac{\pi}{4}

  4. What is dV in spherical coordinates?

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    ρ2sin⁡φ dρ dφ dθ\rho^{2}\sin\varphi\,d\rho\,d\varphi\,d\theta

  5. Find the Jacobian of x = 2u, y = 3v.

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    6

  6. Find the volume of a cylinder of radius 1 and height 4 with cylindrical coordinates.

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    4π

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