Calculus 3, Exams of Calculus

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Calculus 3

January 20, 2022

Chapter 15. Multiple Integrals 15.2. Double Integrals over General Regions—Examples and Proofs of Theorems

Table of contents

• 1 Exercise 15.2.
• 2 Exercise 15.2.
• 3 Exercise 15.2.
• 4 Exercise 15.2.
• 5 Exercise 15.2.

Exercise 15.2.

Exercise 15.2.20. Sketch the region of integration and evaluate the double integral

∫ (^) π

0

∫ (^) sin x

0

y dy dx.

Solution. The region is:

We evaluate the iterated integral as: ∫ (^) π

0

∫ (^) sin x

0

y dy dx =

∫ (^) π

0

y 2 2

y =sin x

y =

dx =

∫ (^) π

0

sin^2 x 2 −^0 dx

Exercise 15.2.

Exercise 15.2.20. Sketch the region of integration and evaluate the double integral

∫ (^) π

0

∫ (^) sin x

0

y dy dx.

Solution. The region is:

We evaluate the iterated integral as: ∫ (^) π

0

∫ (^) sin x

0

y dy dx =

∫ (^) π

0

y 2 2

y =sin x

y =

dx =

∫ (^) π

0

sin^2 x 2 −^0 dx

Exercise 15.2.

Exercise 15.2.40. Sketch the region of integration and write the equivalent double integral with the order of integration reverse: ∫ 2 0

∫ (^4) −y 2

0

y dx dy.

Solution. We first have x ranging from 0 to 4 − y 2 , and second y ranges from 0 to 2. So the region is:

Exercise 15.2.

Exercise 15.2.40. Sketch the region of integration and write the equivalent double integral with the order of integration reverse: ∫ 2 0

∫ (^4) −y 2

0

y dx dy.

Solution. We first have x ranging from 0 to 4 − y 2 , and second y ranges from 0 to 2. So the region is:

Now we can interpret that first y ranges from 0 to the curve x = 4 − y 2 (or y =

4 − x, since y ≥ 0 on the region) and second x ranges from 0 to 4. So the integral becomes ∫ 4 0

∫ √ 4 −x

0

y dy dx.

Exercise 15.2.

Exercise 15.2.50. Sketch the region of integration, reverse the order of integration, and evaluate the integral: ∫ (^2)

0

∫ (^4) −x 2

0

xe^2 y 4 − y dy dx.

Solution. We first have y ranging from 0 to 4 − x^2 , and second x ranges from 0 to 2. So the region is:

Exercise 15.2.

Exercise 15.2.50. Sketch the region of integration, reverse the order of integration, and evaluate the integral: ∫ (^2)

0

∫ (^4) −x 2

0

xe^2 y 4 − y dy dx.

Solution. We first have y ranging from 0 to 4 − x^2 , and second x ranges from 0 to 2. So the region is: Now we can interpret that first x ranges from 0 to the curve y = 4 − x^2 (or x =

4 − y , since x ≥ 0 on the region) and second y ranges from 0 to 4. So the integral becomes ∫ 4 0

∫ √ 4 −y

0

xe^2 y 4 − y dx dy^.

Exercise 15.2.50 (continued)

Exercise 15.2.50. Sketch the region of integration, reverse the order of integration, and evaluate the integral: ∫ (^2)

0

∫ (^4) −x 2

0

xe^2 y 4 − y dy dx. Solution (continued). We now evaluate the new iterated integral: ∫ (^4)

0

∫ √ 4 −y

0

xe^2 y 4 − y dx dy^ =

0

x^2 e^2 y 2(4 − y )

x=√ 4 −y

x=

dy

0

4 − y )^2 e^2 y 2(4 − y )

− 0 dy =

0

(4 − y )e^2 y 2(4 − y )

dy =

0

e^2 y 2

dy

e^2 y 4

y =

y =

e2(4) 4 −^

e2(0) 4 =^

e^8 − 1

Exercise 15.2.

Exercise 15.2.58. Find the volume of the solid that is bounded above the cylinder z = x^2 and below by the region enclosed by the parabola y = 2 − x^2 and the line y = x in the xy -plane. Solution. The region R is:

Exercise 15.2.

Exercise 15.2.58. Find the volume of the solid that is bounded above the cylinder z = x^2 and below by the region enclosed by the parabola y = 2 − x^2 and the line y = x in the xy -plane. Solution. The region R is:

First y ranges from x to 2 − x^2 , and second x ranges from −2 to 1. Since z = f (x, y ) = x^2 is nonnegative over R then the desired volume is

V =

− 2

∫ (^2) −x 2

x

x^2 dy dx.

Exercise 15.2.58 (continued)

Solution (continued). So the volume is

V =

− 2

∫ (^2) −x 2

x

x^2 dy dx =

− 2

x^2 y

∣∣y =2−x 2 y =x dx

− 2

x^2 (2−x^2 )−x^2 (x) dx =

− 2

2 x^2 −x^4 −x^3 dx =

2 x^3 3 −^

x^5 5 −^

x^4 4

x=

x=− 2

=

2(1)^3

5 5

4 4

2(−2)^3

5 5

4 4

3 −^

5 −^

3 −^

60 −^

60 −^

60 −^

Exercise 15.2.

Exercise 15.2.76. (Unbounded Region) Integrate f (x, y ) =

(x^2 − x)(y − 1)^2 /^3

over the infinite rectangle 2 ≤ x < ∞, 0 ≤ y ≤ 2.

Solution. We want to find

2

0

(x^2 − x)(y − 1)^2 /^3

dy dx. This is an improper integral and so we write it as a limit: ∫ (^) ∞

2

0

(x^2 − x)(y − 1)^2 /^3

dy dx = lim b→∞

∫ (^) b

2

0

(x^2 − x)(y − 1)^2 /^3

dy dx

= lim b→∞

∫ (^) b

2

x^2 − x

(y − 1)^1 /^3 1 / 3

y =

y =

dx

= lim b→∞

∫ (^) b

2

x^2 − x

3((2) − 1)^1 /^3 − 1

x^2 − x

3((0) − 1)1/ 3 dx

Exercise 15.2.76 (continued)

Exercise 15.2.76. (Unbounded Region) Integrate f (x, y ) =

(x^2 − x)(y − 1)^2 /^3

over the infinite rectangle 2 ≤ x < ∞, 0 ≤ y ≤ 2. Solution (continued).

= lim b→∞

∫ (^) b

2

x^2 − x

dx = 6 lim b→∞

∫ (^) b

2

x − 1

x

dx by partial fractions

= 6 lim b→∞

(ln(x − 1) − ln x)|xx==2b = 6 lim b→∞

ln

x − 1 x

x=b

x=

= 6 lim b→∞

ln

b − 1 b

− 6 ln

= −6 ln(1/2) = 6 ln 2.