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Multiple Integrals

3.6 Triple Integrals in Cylindrical Coordinates

17 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

Use (r,θ,z)(r,\theta,z) to denote cylindrical coordinates.

  1. Draw r=0r=0.

  2. Draw r=1r=1.

  3. Draw θ=0\theta=0.

  4. Draw θ=π ⁣/4\theta=\nicefrac{\pi}{4}.

Answer

(a), (b)

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

(c), (d)

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Figure from prob_s3.6, line 30

Full solution
Q2Stage 1

Sketch the points with the specified cylindrical coordinates.

  1. r=1r=1, θ=0\theta=0, z=0z=0

  2. r=1r=1, θ=π4\theta=\frac{\pi}{4}, z=0z=0

  3. r=1r=1, θ=π2\theta=\frac{\pi}{2}, z=0z=0

  4. r=0r=0, θ=π\theta=\pi, z=1z=1

  5. r=1r=1, θ=π4\theta=\frac{\pi}{4}, z=1z=1

Answer

Figure from prob_s3.6, line 101

Figure from prob_s3.6, line 101

Full solution
Q3Stage 1

Convert from cylindrical to Cartesian coordinates.

  1. r=1r=1, θ=0\theta=0, z=0z=0

  2. r=1r=1, θ=π4\theta=\frac{\pi}{4}, z=0z=0

  3. r=1r=1, θ=π2\theta=\frac{\pi}{2}, z=0z=0

  4. r=0r=0, θ=π\theta=\pi, z=1z=1

  5. r=1r=1, θ=π4\theta=\frac{\pi}{4}, z=1z=1

Answer

(a) (1,0,0)(1,0,0) (b) (12,12,0)\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right) (c) (0,1,0)(0,1,0) (d) (0,0,1)(0,0,1) (e) (12,12,1)\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},1\right)

Full solution
Q4Stage 1

Convert from Cartesian to cylindrical coordinates.

  1. (1,1,2)(1,1,2)

  2. (−1,−1,2)(-1,-1,2)

  3. (−1,3,0)(-1,\sqrt{3}, 0)

  4. (0,0,1)(0,0,1)

Answer

(a) r=2r=\sqrt{2}, z=2z=2, θ=π4\theta=\frac{\pi}{4} (plus possibly any integer multiple of 2π2\pi)

(b) r=2r=\sqrt{2}, z=2z=2, θ=5π4\theta=\frac{5\pi}{4} (plus possibly any integer multiple of 2π2\pi)

(c) r=2r=2, z=0z=0, θ=2π3\theta=\frac{2\pi}{3} (plus possibly any integer multiple of 2π2\pi)

(d) r=0r=0, z=1z=1, θ=arbitrary\theta=\text{arbitrary}

Full solution
Q5Stage 1

Rewrite the following equations in cylindrical coordinates.

  1. z=2xyz=2xy

  2. x2+y2+z2=1x^2+y^2+z^2=1

  3. (x−1)2+y2=1(x-1)^2 + y^2 =1

Answer

(a) z=r2sin⁡(2θ)z=r^2\sin(2\theta) (b) r2+z2=1r^2+z^2=1 (c) r=2cos⁡θr=2\cos\theta

Full solution

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q6Stage 2

Use cylindrical coordinates to evaluate the volumes of each of the following regions.

  1. Above the xyxy-plane, inside the cone z=2a−x2+y2z=2a-\sqrt{x^2+y^2} and inside the cylinder x2+y2=2ayx^2+y^2=2ay, where a>0a>0 is a constant.

  2. Above the xyxy-plane, under the paraboloid z=1−x2−y2z=1-x^2-y^2 and in the wedge −x≤y≤3x-x\le y\le \sqrt{3}x.

  3. Above the paraboloid z=x2+y2z=x^2+y^2 and below the plane z=2yz=2y.

Answer

(a) a3(2π−329)a^3\big(2\pi-\frac{32}{9}\big) (b) 748π\frac{7}{48}\pi (c) π2\frac{\pi}{2}

Full solution
Q7Stage 2Past exam · M200 2008A

Let E be the region bounded between the parabolic surfaces z=x2+y2z = x^2 + y^2 and z=2−x2−y2z = 2 - x^2 - y^2 and within the cylinder x2+y2≤1x^2 + y^2 \le 1. Calculate the integral of f(x,y,z)=(x2+y2)3/2f(x,y,z) = {(x^2 + y^2)}^{3/2} over the region EE.

Answer

8π35\frac{8\pi}{35}

Full solution
Q8Stage 2Past exam · M200 2010D

Let EE be the region bounded above by the sphere x2+y2+z2=2x^2 + y^2 + z^2 = 2 and below by the paraboloid z=x2+y2z = x^2 + y^2. Find the centroid of EE.

Answer

(0,0,7162−14≈0.811)\big(0,0,\frac{7}{16\sqrt{2}-14} \approx 0.811\big)

Full solution
Q9Stage 2Past exam · M200 2012A

Let EE be the smaller of the two solid regions bounded by the surfaces z=x2+y2z = x^2 + y^2 and x2+y2+z2=6x^2 + y^2 + z^2 = 6. Evaluate ∭E(x2+y2) dV\tripInt_E (x^2+y^2)\ \dee{V} .

Answer

π[4856−32815]≈1.65π\pi\left[\frac{48}{5}\sqrt{6}-\frac{328}{15}\right] \approx 1.65\pi

Full solution
Q10Stage 2Past exam · M200 2013D

Let a>0a > 0 be a fixed positive real number. Consider the solid inside both the cylinder x2+y2=axx^2 + y^2 = ax and the sphere x2+y2+z2=a2x^2 + y^2 + z^2 = a^2. Compute its volume.

You may use that ∫sin⁡3(θ)=112cos⁡(3θ)−34cos⁡(θ)+C\int \sin^3(\theta) =\frac{1}{12}\cos(3\theta) -\frac{3}{4}\cos(\theta) +C

Answer

4a33[π2−23]\frac{4a^3}{3}\left[\frac{\pi}{2} - \frac{2}{3} \right]

Full solution
Q11Stage 2Past exam · M200 2014A

Let EE be the solid lying above the surface z=y2z = y^2 and below the surface z=4−x2z = 4 - x^2. Evaluate

∭Ey2 dV\begin{equation*} \tripInt_E y^2\ \dee{V} \end{equation*}

You may use the half angle formulas:

sin⁡2θ=1−cos⁡(2θ)2,cos⁡2θ=1+cos⁡(2θ)2\begin{equation*} \sin^2\theta = \frac{1-\cos(2\theta)}{2},\qquad \cos^2\theta = \frac{1+\cos(2\theta)}{2} \end{equation*}
Answer

16π3\frac{16\pi}{3}

Full solution
Q12Stage 2

The centre of mass (xˉ,yˉ,zˉ)(\bar x,\bar y, \bar z) of a body BB having density ρ(x,y,z)\rho(x,y,z) (units of mass per unit volume) at (x,y,z)(x,y,z) is defined to be

xˉ=1M∭Bxρ(x,y,z) dVyˉ=1M∭Byρ(x,y,z) dVzˉ=1M∭Bzρ(x,y,z) dV\begin{equation*} \bar x=\frac{1}{M}\tripInt_B x\rho(x,y,z)\ \dee{V}\quad \bar y=\frac{1}{M}\tripInt_B y\rho(x,y,z)\ \dee{V}\quad \bar z=\frac{1}{M}\tripInt_B z\rho(x,y,z)\ \dee{V} \end{equation*}

where

M=∭Bρ(x,y,z) dV\begin{equation*} M=\tripInt_B \rho(x,y,z)\ \dee{V} \end{equation*}

is the mass of the body. So, for example, xˉ\bar x is the weighted average of xx over the body. Find the centre of mass of the part of the solid ball x2+y2+z2≤a2x^2+y^2+z^2\le a^2 with x≥0x\ge 0, y≥0y\ge 0 and z≥0z\ge 0, assuming that the density ρ\rho is constant.

Hint

Use cylindrical coordinates.

Answer

xˉ=yˉ=zˉ=38a\bar x=\bar y=\bar z=\frac{3}{8}a

Full solution
Q13Stage 2Past exam · M200 2003D

A sphere of radius 2m2{\rm m} centred on the origin has variable density 53(z2+1)\frac{5}{\sqrt{3}}(z^2+1)kg/m3{\rm m}^3. A hole of diameter 1m is drilled through the sphere along the zz–axis.

  1. Set up a triple integral in cylindrical coordinates giving the mass of the sphere after the hole has been drilled.

  2. Evaluate this integral.

Answer

(a) mass=∫1/22dr∫−4−r24−r2dz∫02πdθ 53(z2+1)r\dst\text{mass} = \int_{1/2}^2 \dee{r} \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} \dee{z}\int_0^{2\pi}\dee{\theta}\ \frac{5}{\sqrt{3}}(z^2+1)r

(b) 525245π≈153.7kg\frac{525}{24}\sqrt{5}\pi\approx 153.7\text{kg}

Full solution
Q14Stage 2Past exam · M200 2001A

Consider the finite solid bounded by the three surfaces: z=e−x2−y2z=e^{-x^2-y^2}, z=0z=0 and x2+y2=4x^2+y^2=4.

  1. Set up (but do not evaluate) a triple integral in rectangular coordinates that describes the volume of the solid.

  2. Calculate the volume of the solid using any method.

Answer

(a) ∫−22dx∫−4−x24−x2dy∫0e−x2−y2dz\dst\int_{-2}^2 \dee{x} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\dee{y}\int_0^{e^{-x^2-y^2}}\dee{z} (b) π[1−e−4]≈3.084\pi\big[1-e^{-4}\big]\approx 3.084

Full solution
Q15Stage 2Past exam · M200 2000A

Find the volume of the solid which is inside x2+y2=4x^2 + y^2 = 4, above z=0z = 0 and below 2z=y2z = y.

Answer

83\frac{8}{3}

Full solution

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q16Stage 3Past exam · M200 2014D

The density of hydrogen gas in a region of space is given by the formula

ρ(x,y,z)=z+2x21+x2+y2\begin{equation*} \rho(x,y,z) =\frac{z+2x^2}{1+x^2+y^2} \end{equation*}

  1. At (1,0,−1)(1,0,-1), in which direction is the density of hydrogen increasing most rapidly?

  2. You are in a spacecraft at the origin. Suppose the spacecraft flies in the direction of <0,0,1>\llt 0,0,1\rgt. It has a disc of radius 11, centred on the spacecraft and deployed perpendicular to the direction of travel, to catch hydrogen. How much hydrogen has been collected by the time that the spacecraft has traveled a distance 22?

    You may use the fact that ∫02πcos⁡2θ dθ=π\int_0^{2\pi}\cos^2\theta\ \dee{\theta} =\pi.

Answer

(a) The unit vector in the direction of maximum rate of increse is 110(3,0,1)\frac{1}{\sqrt{10}}(3,0,1).

(b) 2π2\pi

Full solution
Q17Stage 3

A torus of mass MM is generated by rotating a circle of radius aa about an axis in its plane at distance bb from the centre (b>a)(b>a). The torus has constant density. Find the moment of inertia about the axis of rotation. By definition the moment of intertia is ∭r2dm\tripInt r^2 \dee{m} where dm\dee{m} is the mass of an infinitesmal piece of the solid and rr is its distance from the axis.

Answer

M(34a2+b2)M\big(\frac{3}{4}a^2+b^2\big)

Full solution

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From the UBC Math 200 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.