APPM 2350—Exam 3 (Chapter 12 and Scalar Line Integrals and Scalar Surface Integrals )Wednesday April 15th, 7pm-8:30pm 2020

For each problem below, provide work justifying your full set-up of the limits of integration and fully sim-plify integrand(s) so all that is left to do is integrate. For full credit, set-up using the coordinate system thatleads to the simplest integral(s).

Problem 1 (20 points) Consider a thin wire in the xy-plane that lies along the boundary of the region enclosedby the two curves

y = 2− x and x = 4− y2 ,where the units of x and y are length. The density along the wire, with units of mass per unit length, is givenby

δ(x, y) = αxy2 ,

where α is a known constant. Fully set-up integral(s) to find the total mass, M , of the wire. DO NOTEVALUATE your resulting integral(s).SOLUTION:

First to find the intercepts of these curves:

2− y = 4− y2

→ y2 − y − 2 = 0

→ (y + 1)(y − 2) = 0

→ y = 2,−1

→ (x, y) = (0, 2) and (3,−1)

Now parameterize each separate line segment. Simplest parameterizations are:

C1 : ~r1(t) = 〈t, 2− t〉 for t ∈ [0, 3]

C2 : ~r2(t) = 〈4− t2, t〉 for t ∈ [−1, 2]

Find the Jacobian for each line segment:

C1 : ~v1(t) = 〈1,−1〉 =⇒ ||~v1(t)|| =√

2

C2 : ~v2(t) = 〈−2t, 1〉 =⇒ ||~v1(t)|| =√

4t2 + 1

The mass of the wire along C = C1 ∪ C2 is then

M =

∫Cδ(x, y)ds

=

∫C1

δ(x(t), y(t))||~v1(t)||dt+

∫C2

δ(x(t), y(t))||~v2(t)||dt

=

∫ 3

0

√2αt(2− t)2 dt+

∫ 2

−1α(4− t2)t2

√1 + 4t2 dt

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Problem 2 (20 points) Consider the integral

∫ 3

1

∫ 4−x

1

(x− yx+ y

)dy dx .

Use the transformation

u = x− y and v = x+ y

to fully set-up the equivalent integral(s) over an appropriate region in the uv-plane. DO NOT EVALUATE.For full credit, choose the set-up that results in the fewest possible number of integrals. Pro tip: For yourown use (it won’t be graded) you might consider drawing the region of integration in the xy-plane and in thenew uv-plane.SOLUTION:

Based on the original limits of integration, the region of integration in the xy-plane is a triangle bounded onthe left by x = 1, on the bottom by y = 1, and on the top (or right) by x+ y = 4.The suggested transformation, u = x − y and v = x + y, can be inverted by adding these two equation toyield u+ v = 2x, then subtracting them to yield u− v = −2y, from which we can find

x =u+ v

2and y =

v − u2

.

The Jacobian is then

J(u, v) = xu yv − yu xv =1

4+

1

4= 1/2 .

The integrand itself becomes,

(x− yx+ y

)=u

v.

The boundaries in the original xy-plane transform as follows:

Thus, the final integral becomes ∫ 4

v=2

∫ v−2

u=2−v

(uv

) ∣∣∣∣12∣∣∣∣ du dv .

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Problem 3 (20 points) Let E be the solid region that consists of all points such that

x2 + y2 + z2 ≥ 4 AND z ≤ 6 AND z ≥√

3x2 + 3y2 .

To help you out, here is a sketch of the cross section of the surfaces of interest in an arbitrary θ plane. Thefigure is drawn to scale, but you will need to find the specific intersection locations, and most importantly,correctly identify the region of integration. Pro tip: Double check your intersections before proceeding withthe problem! And then check them one more time.

r axis

z axis

The temperature in the solid is given by

T (x, y, z) = z(x2 + y2) .

Fully set up, BUT DO NOT EVALUATE, integral(s) to find

(a) The average temperature of the region using spherical coordinates in the order dρ, dφ, dθ

(b) The volume of the region using cylindrical coordinates in the order dr, dz, dθSOLUTION:

(a) The average temperature of the region is

∫∫∫E

T (x, y, z) dV∫∫∫E

dV. First set up the bounds of E in spheri-

cal:

x2 + y2 + z2 = 4→ ρ = 2

z = 6→ ρ =6

cos(φ)= 6 sec(φ)

z =√

3x2 + 3y2 → ρ cos(φ) =

√3ρ2 sin2(φ) cos2(θ) + 3ρ2 sin2(φ) sin2(θ)

→ cos(φ) =√

3 sin(φ)→ φ =π

6

The temperature in spherical coordinates is T (ρ, φ, θ) = ρ3 cos(φ) sin2(φ). The average temperatureis then

Tave =

∫ 2π0

∫ π60

∫ 6 sec(φ)2 ρ5 cos(φ) sin3(φ) dρ dφ dθ∫ 2π

0

∫ π60

∫ 6 sec(φ)2 ρ2 sin(φ) dρ dφ dθ

(b) We need two separate triple integrals to set-up this volume using cylindrical coordinates and theordering of integration specified:

First the intercept of r2 + z2 = 4 and z =√

3r needs to be found.

r2 +(√

3r)2

= 4→ 4r2 = 4

→ r = 1→ z =√

3

Then the volume is given by:

V =

∫ 2π

0

∫ 2

√3

∫ z√3

√4−z2

r drdzdθ +

∫ 2π

0

∫ 6

2

∫ z√3

0r drdzdθ

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Problem 4 (20 points) Consider the hemispherical surface given by x2 + y2 + z2 = 9 for x ≥ 0. Set upintegral(s) to find x̄ (the x-component of the centroid) of the part of this hemispherical surface that lies insidethe cylinder y2+z2 = 4. For full credit, set this up using the coordinate system that leads to the simplest andfewest number of integral(s). Pro tip: sketch the object of interest, and reorient it and the axes as necessary,until you have it clear in your head what it looks like.SOLUTION:

To determine x̄, we need to calculate

x̄ =

∫∫S x dS∫∫S dS

.

where the surface S is the portion of the spherical surface inside the cylinder y2 + z2 = 4.

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Problem 5 (20 points) Consider a cone of height H whose base has radius R. At the vertex of the cone(located at x = 0, y = 0, z = H) there is the source of a charge. At the base of the cone is a circular metaldisk in the z = 0 plane. On that circular disk, the charge density at each point P is given by

δ(x, y) =α

d(x, y),

where d(x, y) is the distance from P to the charge source location, and α is a known constant. Set up, butDO NOT EVALUATE integral(s), to find the total charge on the circular metal disk. (Pro Tip: Read thatprevious sentence again until you are confident about what you are integrating). For full credit, set up yourcalculation using the coordinate system, and integration order, that leads to the simplest integral(s).SOLUTION:

This problem is best worked in polar coordinates where the region of integration is the circular disk of radiusR in the xy plane, x2 + y2 = R2.The distance from the charge to an arbitrary location, (x, y), on the plate is

d(x, y) =α√

x2 + y2 +H2,

or in polar,d(r, θ) =

α√r2 +H2

.

Converting everything into polar coordinates, and integrating δ over the circular region, leads to a total chargeof ∫ 2π

θ=0

∫ R

r=0

α√r2 +H2

r dr dθ .

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End Of Exam