Download Double Integrals over Rectangular Regions: Math 210 and more Lecture notes Mathematics in PDF only on Docsity!
Math 210 (Lesieutre) 13.1: Double integrals over rectangular regions March 3, 2017
Problem 1. Compute the following double integrals. Sketch the region R over which the integral is being taken.
a)
0
1
xy dy dx
The inner integral is everything from the second integral sign to the dwhatever. ∫ (^2)
1
xy dy = y^2 2 x
∣∣^2
y=
x −
x =
x.
The outer integral is then ∫ (^1)
0
x dx =
x^2 2
∣∣^1
x=
That’s our final answer.
b)
1
0
xy dx dy
Inner: (^) ∫ (^1)
0
xy dx = y x^2 2
1 x=
y 2
Outer: (^) ∫ (^2)
1
y 2 dy = y^2 4
2 1
Same answer as before. Coincidence? No – Fubini’s theorem. Changing the order of the variables doesn’t change the value of the integral.
Problem 2. a)
0
0
yexy^ dy dx
The inner integral is
0
yexy^ dy, which is moderately unpleasant, since we have to integrate
by parts. If we change the order of integration, it’s a little better: ∫ (^1)
0
0
yexy^ dx dy
The inner integral is given by ∫ (^2)
0
yexy^ dx = exy
∣∣^2
x= = e^2 y^ − e^0 y^ = e^2 y^ − 1.
The final answer is ∫ (^1)
0
e^2 y^ − 1 dy =
e^2 y^ − y
1
y=
e^2 − 1
e^2 −
Notice that if we tried to do it in the other order, it would be tough going:
yexy^ dy would require an integration by parts. In some cases, it’s even worse: the integral simply can’t be done unless the order is right.
b) What is the average value of f (x, y) = yexy^ on the region R? It’s given by 1 area(R)
0
0
yexy^ dy dx =
e^2 −
e^2 −
Problem 3. a) Sketch the region of integration for
∫ (^2)
0
∫ (^) x 2
0
y dy dx.
It’s the region between y = 0 and y = x^2 with 0 ≤ x ≤ 2.
b) Evaluate the integral. Inner: (^) ∫ x^2 0
y dy = y^2 2
∣∣x^2 0
x^4 2
Outer: (^) ∫ (^2)
0
x^4 2 dx = x^5 10
2 0
c) Rewrite the integral with the variables in the opposite order. This time y is going to go on the outside. Based on the picture, we need to go from y = 0 to y = 4. For a given value of y, what x’s do we want? It’s frmo x = 0 to
y to 2. ∫ (^4)
0
√y^ y dx dy.
d) Evaluate the integral. Inner: (^) ∫ (^2)
√y^ y dx^ =^ y(2^ −
y) = 2y − y^3 /^2.
Outer: (^) ∫ (^4)
0
2 y − y^3 /^2 dy =
y^2 −
y^5 /^2
4
0
