Math Buddyv1.0

Triple integral calculator with steps

Type or scan a triple integral. MathBuddy integrates one variable at a time, in rectangular, cylindrical or spherical coordinates.

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What this page covers

  • Triple integrals with numeric limits over a box, integrated from the inside out.
  • Limits that depend on the outer variables, for solids such as a tetrahedron.
  • Cylindrical and spherical coordinates, with the extra factor rr or ρ2sin⁡ϕ\rho^2 \sin\phi written out.
  • Not covered here: single integrals (use Integral calculator with steps) and double integrals over a region in the plane (use Double integral calculator).

How to enter the problem

  • Type three integral signs with their limits, then the function and the order, such as dz dy dx. The innermost limits go with the first differential.
  • Scan it. Fit the one problem in the frame.
  • Choose a photo. Several problems in one photo are solved as a list.
  • Then tap Solve.

Worked examples

Example 1

Problem

Find ∫01∫0x∫0yx dz dy dx\displaystyle \int_0^1 \int_0^x \int_0^y x\,dz\,dy\,dx.

Answer

Verified

The answer is

18\frac{1}{8}

Explanation

  1. Integrate with respect to zz
    xx does not depend on zz, so ∫0yx dz=xy\int_0^y x\,dz = xy.
  2. Integrate with respect to yy
    ∫0xxy dy=x⋅x22=x32\int_0^x xy\,dy = x \cdot \frac{x^2}{2} = \frac{x^3}{2}
  3. Integrate with respect to xx
    ∫01x32 dx=12⋅14=18\int_0^1 \frac{x^3}{2}\,dx = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}

Example 2

Problem

Find ∭Ex dV\displaystyle \iiint_E x\,dV, where E\displaystyle E is the quarter cylinder x2+y2≤4\displaystyle x^2 + y^2 \le 4, x≥0\displaystyle x \ge 0, y≥0\displaystyle y \ge 0, 0≤z≤3\displaystyle 0 \le z \le 3.

Answer

Verified

The answer is

88

Explanation

  1. Describe the solid in cylindrical coordinates
    0≤r≤20 \le r \le 2, 0≤θ≤π20 \le \theta \le \frac{\pi}{2}, 0≤z≤30 \le z \le 3, with x=rcos⁡θx = r\cos\theta and dV=r dr dθ dzdV = r\,dr\,d\theta\,dz.
  2. Write the integral
    ∫03∫0π/2∫02r2cos⁡θ dr dθ dz\int_0^3 \int_0^{\pi/2} \int_0^2 r^2 \cos\theta\,dr\,d\theta\,dz. The integrand rcos⁡θr\cos\theta picks up one more factor rr from dVdV.
  3. Integrate with respect to rr
    ∫02r2 dr=83\int_0^2 r^2\,dr = \frac{8}{3}, which leaves 83cos⁡θ\frac{8}{3}\cos\theta.
  4. Integrate with respect to θ\theta
    ∫0π/283cos⁡θ dθ=83\int_0^{\pi/2} \frac{8}{3}\cos\theta\,d\theta = \frac{8}{3}
  5. Integrate with respect to zz
    ∫0383 dz=8\int_0^3 \frac{8}{3}\,dz = 8

Example 3

Problem

Use spherical coordinates to find ∭Exz dV\displaystyle \iiint_E xz\,dV, where E\displaystyle E is the part of the ball x2+y2+z2≤4\displaystyle x^2 + y^2 + z^2 \le 4 with x,y,z≥0\displaystyle x, y, z \ge 0.

Answer

Verified

The answer is

3215\frac{32}{15}

Explanation

  1. Describe the solid in spherical coordinates
    0≤ρ≤20 \le \rho \le 2, 0≤ϕ≤π20 \le \phi \le \frac{\pi}{2}, 0≤θ≤π20 \le \theta \le \frac{\pi}{2}, with dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2 \sin\phi\,d\rho\,d\phi\,d\theta.
  2. Rewrite the integrand
    x=ρsin⁡ϕcos⁡θx = \rho\sin\phi\cos\theta and z=ρcos⁡ϕz = \rho\cos\phi, so xz dV=ρ4sin⁡2ϕcos⁡ϕcos⁡θ dρ dϕ dθxz\,dV = \rho^4 \sin^2\phi\cos\phi\cos\theta\,d\rho\,d\phi\,d\theta.
  3. Separate the three integrals
    The limits are constants and the integrand is a product, so the integral is ∫02ρ4 dρ⋅∫0π/2sin⁡2ϕcos⁡ϕ dϕ⋅∫0π/2cos⁡θ dθ\int_0^2 \rho^4\,d\rho \cdot \int_0^{\pi/2} \sin^2\phi\cos\phi\,d\phi \cdot \int_0^{\pi/2} \cos\theta\,d\theta.
  4. Evaluate each one
    ∫02ρ4 dρ=325\int_0^2 \rho^4\,d\rho = \frac{32}{5}, ∫0π/2sin⁡2ϕcos⁡ϕ dϕ=13\int_0^{\pi/2} \sin^2\phi\cos\phi\,d\phi = \frac{1}{3} (substitute u=sin⁡ϕu = \sin\phi), and ∫0π/2cos⁡θ dθ=1\int_0^{\pi/2} \cos\theta\,d\theta = 1.
  5. Multiply
    325⋅13⋅1=3215\frac{32}{5} \cdot \frac{1}{3} \cdot 1 = \frac{32}{15}

Common mistakes

  • Leaving out ρ2sin⁡ϕ\rho^2 \sin\phi in spherical coordinates. In worked example 3 that gives 43\frac{4}{3} instead of 3215\frac{32}{15}.
  • Leaving out the factor rr in cylindrical coordinates: dV=r dr dθ dzdV = r\,dr\,d\theta\,dz, not dr dθ dzdr\,d\theta\,dz.
  • Matching limits to the wrong variable. The innermost limits belong to the first differential, and only they may contain the outer variables.
  • Mixing up ϕ\phi and θ\theta. Here ϕ\phi is measured down from the positive zz-axis and runs from 0 to at most π\pi; θ\theta turns around the zz-axis.

Checks, assumptions and limits

  • Check a box integral by splitting it. In the runnable example xyzxyz separates, so the answer is ∫01x dx⋅∫02y dy⋅∫03z dz=12⋅2⋅92=92\int_0^1 x\,dx \cdot \int_0^2 y\,dy \cdot \int_0^3 z\,dz = \frac{1}{2} \cdot 2 \cdot \frac{9}{2} = \frac{9}{2}.
  • Check the setup with the integrand 1. The limits of worked example 2 then give 14π⋅22⋅3=3π\frac{1}{4}\pi \cdot 2^2 \cdot 3 = 3\pi, the volume of that quarter cylinder.
  • Angles are in radians. Limits such as π2\frac{\pi}{2} for θ\theta and ϕ\phi assume radians.
  • MathBuddy can make mistakes. Double check important steps.

Related tasks

Frequently asked questions

When the solid is a ball, part of a ball or a cone with its tip at the origin, or the integrand contains x2+y2+z2x^2 + y^2 + z^2. Then ρ\rho often has constant limits. Remember the volume element ρ2sin⁡ϕ dρ dϕ dθ\rho^2 \sin\phi\,d\rho\,d\phi\,d\theta.

For a box with constant limits and a continuous function, every order gives the same value. For other solids, changing the order means rewriting the limits, and one order is often much easier than another.

Yes. Use the camera on the one problem, or choose a photo. Several problems in one photo are solved as a list.

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