MA 243 Calculus III Fall 2017 Exam III - Solutions

1. (20 points) On Exam 2, you expressed the volume under z = 1− 12

(x2 + y2

)and above the xy plane as a

double integral in rectangular coordinates. This time, express this volume as an integral in polar coordinatesand calculate it.

Vol =

∫ √2

0

∫ 2π

0

(1− 1

2r2)

r dθ dr = 2π

∫ √2

0

(r − 1

2r3)

dr = 2π

[1

2r2 − 1

8r4]√2

0

= π

2. (20 points) Use a double integral to calculate the surface area of the portion of the paraboloidz = 1− 1

2

(x2 + y2

)above the xy plane.

Let R be the circle in the xy plane described by the equation x2 + y2 = 2. The surface area is∫∫R

√1 + z2x + z2y dA =

∫∫R

√1 + x2 + y2 dA. This integral is most easily done in polar coordinates.

SA =

∫ √2

0

∫ 2π

0

(1 + r2

)1/2r dθ dr = 2π

∫ √2

0

(1 + r2

)1/2r dr =

2π

3

(3√3− 1

)3. (20 points) Let T be the region inside the triangle with vertices (0, 0, 0), (1, 0, 0) and (1, 3, 0). T hasan area of 3

2 . Use double integration to calculate the y-coordinate of the centroid of T .

y =1

Area(T )

∫ ∫Ty dA =

2

3

∫ 1

0

∫ 3x

0

y dy dx =2

3

∫ 1

0

9

2x2 dx = 1

4. (20 points) Let V be the three dimensional region in the first octant that is above the cone z =√x2 + y2

and below the hemisphere z =√2− x2 − y2

Express the volume of V as a triple integral in:

a) rectangular coordinates b) spherical coordinates

You do not have to calculate any antiderivatives. I am only interested in the triple integral set-up with thecorrect limits of integration.

Volume =

∫ 1

0

∫ √1−x2

0

∫ √2−x2−y2

√x2+y2

1 dz dy dx =

∫ π/4

0

∫ π/2

0

∫ √2

0

ρ2 sinϕdρ dθ dϕ

5. (20 points) Let T be the triangle described in problem 3. Let Ω be the region above T but below theplane z = 3(1− x).

Fill in the limits of integration for each of the following triple integrals so that they each represent the volumeof Ω. You do not have to calculate any antiderivatives. I am only interested in the triple integral set-up withthe correct limits.

Vol(Ω) =

∫ 1

0

∫ 3x

0

∫ 3(1−x)

0

1 dz dy dx =

∫ 3

0

∫ 3−y

0

∫ 1−z/3

y/3

1 dx dz dy