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16 MULTIPLE INTEGRALS

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CHAPTER <strong>16</strong><br />

<strong>MULTIPLE</strong> <strong>INTEGRALS</strong><br />

• Present some different uses of polar integrals:<br />

1. Compute ∫∫ (<br />

R x 2 + y 2) 2 dA,whereR is the region enclosed by x 2 + y 2 = 4 between the lines y = x<br />

and y =−x.<br />

y<br />

2<br />

_2<br />

_1<br />

0 1<br />

2<br />

x<br />

_2<br />

2. Compute the volume between the cone z 2 = x 2 + y 2 and the paraboloid z = 4 − x 2 − y 2 .<br />

GROUPWORK1:ThePolarAreaFormula<br />

Perhaps start by having students guess the final answer to Problem 1. Notice that the students will have to be<br />

able to antidifferentiate cos 2 θ and sin 2 θ to complete this activity.<br />

Answers:<br />

1. 2 ∫ π<br />

0 1 2 (1 + cos θ)2 dθ = 3π 2<br />

2. ∫ π<br />

0<br />

[<br />

12<br />

(1 + sin θ) 2 − 1 2 (1)2] dθ = π 4 + 2<br />

GROUP WORK 2: Fun with Polar Area<br />

Many students will write ∫ 3π ∫ θ<br />

0 0<br />

rdrdθ. Try to get them to see for themselves that they must now subtract<br />

the area of the region that is counted twice in this expression.<br />

Answers:<br />

1. 9 2 π3 − 19<br />

6 π3 = 4 3 π3 2. ∫ π<br />

0<br />

∫ 2<br />

0 e−r 2 rdrdθ = π (<br />

2 1 − e<br />

−4 )<br />

GROUP WORK 3: Fun with Polar Volume<br />

If the students get stuck on Problem 2, point out that the intersection can be shown to be x 2 + y 2 = 3 4 , z = 1 2 ,<br />

and hence the integral is over the region x 2 + y 2 ≤ 3 4<br />

. If time is an issue, the students can be instructed to set<br />

up the integrals without solving them.<br />

Answers:<br />

1. ∫ 2π<br />

0<br />

2. ∫ 2π<br />

0<br />

∫ 1<br />

(√<br />

0 1 − r 2 + 1)<br />

rdrdθ = 5π 3<br />

∫ 3/4<br />

[√<br />

0 1 − r 2 −<br />

(1 − √ )]<br />

(<br />

1 − r 2 rdrdθ =<br />

48<br />

π 37 − 7 √ )<br />

7<br />

894

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