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Step 18: Multivariable calculus

Double and triple integrals

Step 18 · Multivariable calculus topics

Integrate functions over rectangles, more general regions and solids by doing one integral inside another, and use them for volumes and totals.

Double integrals over rectangles

A double integral adds f(x, y) dA over tiny patches of area. When f ≥ 0 it is the volume under the surface z = f(x, y). Compute it as an iterated integral: integrate in one variable with the other held fixed, then integrate the result.

∬Rf(x,y) dA=∫ab∫cdf(x,y) dy dx\iint_R f(x, y)\,dA = \int_{a}^{b}\int_{c}^{d} f(x, y)\,dy\,dx

Example: A rectangle

Find ∫02∫03(x+2y) dy dx\int_{0}^{2}\int_{0}^{3} (x + 2y)\,dy\,dx.

  1. Inside, with x fixed: xy + y² from 0 to 3 gives 3x + 9.
  2. Outside: the integral of 3x + 9 from 0 to 2 is 6 + 18.

Answer24

Example: A volume

Find the volume under z = 4 − x − y above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.

  1. Inside: 4y − xy − y²/2 from 0 to 1 gives 3.5 − x.
  2. Outside: the integral of 3.5 − x from 0 to 1 is 3.5 − 0.5.

Answer3

General regions and order of integration

For a region between two curves, the inner limits are functions. If the region lies between y = g₁(x) and y = g₂(x) for a ≤ x ≤ b, integrate in y first. Swapping the order means describing the same region the other way round.

The triangle 0 ≤ y ≤ x, 0 ≤ x ≤ 2. Integrating in y first, y runs from 0 up to x. In x first, x runs from y across to 2.

Example: A triangle

Find ∬Txy dA\iint_T xy\,dA over the triangle 0 ≤ y ≤ x, 0 ≤ x ≤ 2.

  1. ∫02∫0xxy dy dx=∫02x32 dx\int_{0}^{2}\int_{0}^{x} xy\,dy\,dx = \int_{0}^{2} \frac{x^{3}}{2}\,dx.
  2. That is x⁴ ÷ 8 from 0 to 2.

Answer2

Example: Swapping the order

Rewrite ∫01∫x1ey2 dy dx\int_{0}^{1}\int_{x}^{1} e^{y^{2}}\,dy\,dx and evaluate it.

  1. The region is x ≤ y ≤ 1 for 0 ≤ x ≤ 1, which is also 0 ≤ x ≤ y for 0 ≤ y ≤ 1.
  2. So it equals ∫01∫0yey2 dx dy=∫01y ey2 dy\int_{0}^{1}\int_{0}^{y} e^{y^{2}}\,dx\,dy = \int_{0}^{1} y\,e^{y^{2}}\,dy.
  3. That is 12ey2\frac{1}{2}e^{y^{2}} from 0 to 1.

Answere−12\frac{e - 1}{2}, about 0.86

Triple integrals

A triple integral adds f(x, y, z) dV over a solid. With f = 1 it gives the volume; with f a density it gives the mass. Set the limits from the inside out.

Example: A box

Find ∭Bxyz dV\iiint_B xyz\,dV over the box 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 3.

  1. The integrand splits into a product, so the integral does too.
  2. The three single integrals are 1/2, 2 and 9/2.

Answer4.5

Example: A volume

Find the volume of the solid under the plane x + y + z = 1 with x, y and z all positive.

  1. ∫01∫01−x∫01−x−ydz dy dx\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y} dz\,dy\,dx.
  2. The inner two integrals give (1 − x)² ÷ 2.
  3. The last integral gives 1/6.

Answer16\frac{1}{6}

Try it yourself

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

  1. Find ∫01∫026xy dy dx\int_{0}^{1}\int_{0}^{2} 6xy\,dy\,dx.

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    6

  2. Find the area of the region 0 ≤ y ≤ x², 0 ≤ x ≤ 3 with a double integral.

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    9

  3. Find ∫01∫0x2 dy dx\int_{0}^{1}\int_{0}^{x} 2\,dy\,dx.

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    1

  4. Swap the order of ∫02∫0xf dy dx\int_{0}^{2}\int_{0}^{x} f\,dy\,dx.

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    ∫02∫y2f dx dy\int_{0}^{2}\int_{y}^{2} f\,dx\,dy

  5. Find the volume of the box 0 ≤ x ≤ 2, 0 ≤ y ≤ 3, 0 ≤ z ≤ 4 with a triple integral.

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    24

  6. A plate covers the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 with density x + y. What is its mass?

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    1

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