Triple Integrals in Cylindrical Coordinates

Introduction

Now we finally get to use what we covered about cylindrical coordinates.

Cylindrical coordinates diagram

Recall that x=rcos⁡θy=rsin⁡θz=zr=x2+y2θ=tan⁡−1yxz=z\begin{aligned} x &= r \cos \theta\ y &= r \sin \theta\ z &= z\ \ r &= \sqrt{x^2+y^2}\ \theta &= \tan^{-1} \dfrac{y}{x}\ z &= z \end{aligned}

Therefore, a triple integral in rectangular coordinates can be rewritten (similar to a double integral in polar coordinates): ∭Df(x,y,z) dV=∭Df(rcos⁡θ,rsin⁡θ,z) dz r dr dθ\iiint_D f(x,y,z)\ dV = \iiint_D f(r \cos \theta, r \sin \theta, z)\ dz\ r\ dr\ d\theta

Note carefully the rr in the midst of the differentials.

As with double integrals and polar coordinates, we'll tend to use cylindrical coordinates when we encounter a triple integral with x2+y2x^2+y^2 somewhere.

Examples

Convert the following integral to cylindrical coordinates and evaluate. ∫022∫−8−x28−x2∫−121+x2+y2 dz dy dx\int_0^{2\sqrt{2}} \int_{-\sqrt{8-x^2}}^{\sqrt{8-x^2}} \int_{-1}^2 \sqrt{1+x^2+y^2}\ dz\ dy\ dx

Solution

Half cylinder

Note that the limits on zz will remain the same, rr will go from 00 to 8=22\sqrt{8} = 2\sqrt{2}, and θθ will go from −π/2−π/2 to π/2π/2.

∫022∫−8−x28−x2∫−121+x2+y2 dz dy dx=∫−π/2π/2∫022∫−121+r2⋅r dz dr dθ=∫−π/2π/2∫02231+r2⋅r dr dθ=∫−π/2π/2(1+r2)3/2∣022 dθ=∫−π/2π/226 dθ=26π\begin{aligned} \int_0^{2\sqrt{2}} \int_{-\sqrt{8-x^2}}^{\sqrt{8-x^2}} \int_{-1}^2 &\sqrt{1+x^2+y^2}\ dz\ dy\ dx\ &= \int_{-\pi/2}^{\pi/2} \int_0^{2\sqrt{2}} \int_{-1}^2 \sqrt{1+r^2} \cdot r\ dz\ dr\ d\theta\ &= \int_{-\pi/2}^{\pi/2} \int_0^{2\sqrt{2}} 3\sqrt{1+r^2} \cdot r\ dr\ d\theta\ &= \int_{-\pi/2}^{\pi/2} (1+r^2)^{3/2} \bigg|0^{2\sqrt{2}}\ d\theta\ &= \int{-\pi/2}^{\pi/2} 26\ d\theta = \ans{26\pi} \end{aligned}

Find the volume of the solid DD bounded by the paraboloid z=4−x2−y2z=4−x^2−y^2 and the plane z=0z=0.

Solution

V=∭Dr dz dr dθ=∫02π∫02∫04−r2r dz dr dθ=∫02π∫024r−r3 dr dθ=∫02π2r2−14r4∣02 dθ=∫02π4 dθ=8π\begin{aligned} V &= \iiint_D r\ dz\ dr\ d\theta\ &= \int_0^{2\pi} \int_0^2 \int_0^{4-r^2} r\ dz\ dr\ d\theta\ &= \int_0^{2\pi} \int_0^2 4r-r^3\ dr\ d\theta\ &= \int_0^{2\pi} 2r^2-\dfrac{1}{4}r^4 \bigg|_0^2 \ d\theta\ &= \int_0^{2\pi} 4 \ d\theta = \ans{8\pi} \end{aligned}

  1. Use cylindrical coordinates to evaluate ∫02π∫01∫−11r dz dr dθ\displaystyle\int_{0}^{2\pi} \displaystyle\int_{0}^{1} \displaystyle\int_{-1}^{1} r\ dz\ dr\ d\theta.

    2. 2π2\pi

  2. Use cylindrical coordinates to evaluate ∫−11∫−1−y21−y2∫−11(x2+y2)3/2 dz dx dy\displaystyle\int_{-1}^{1} \displaystyle\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \displaystyle\int_{-1}^{1} (x^2+y^2)^{3/2}\ dz\ dx\ dy.

    3. 4π5\dfrac{4\pi}{5}

  3. Use cylindrical coordinates to evaluate ∫−44∫−16−x216−x2∫x2+y24dz dy dx\displaystyle\int_{-4}^{4} \displaystyle\int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} \displaystyle\int_{\sqrt{x^2+y^2}}^{4} dz\ dy\ dx.

    4. −320π3-\dfrac{320\pi}{3}

  4. Use cylindrical coordinates to find the volume of the solid bounded by the plane z=0z=0 and the hyperboloid z=17−1+x2+y2z=\sqrt{17}-\sqrt{1+x^2+y^2}

    5. 14π3⋅17+2π3\dfrac{14\pi}{3}\cdot\sqrt{17}+\dfrac{2\pi}{3}

  5. Use cylindrical coordinates to find the volume of the solid bounded by the plane z=29z=\sqrt{29} and the hyperboloid z=4+x2+y2.z=\sqrt{4+x^2+y^2}.

    6. (17296+83)2π\left(\dfrac{17\sqrt{29}}{6}+\dfrac{8}{3}\right)2\pi

  6. Use cylindrical coordinates to find the volume of the cylinder in the first octant bounded by the cylinder r=1r=1 and the planes z=xz=x and z=0z=0.

    7. 23\dfrac{2}{3}