# MATA33H3 Lecture Notes - Multiple Integral, Farad, Elementary Function

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Reversing the Order of Integration

Let z=f(x, y) and let Bbe a subset of the Xy-plane. The volume of the ﬁgure

lying below the graph of f(x, y) and over the set Bis denoted RBf(x, y)dA To compute

it requires a double integral. If Bis the region between the curves y=g(x) (below) and

y=h(x) (above) for xbetween aand b, then RBf(x, y)dA =Rb

aRh(x)

g(x)f(x, y)dy dx. Note

that the limits of integration depend only the region Band are independent of the function

z=f(x, y) being integrated. Since the preceding discussion is symmetrical with respect to

xand y, we cab reverse the roles of xand yif convenient. That is, if Bis written instead

as the region bounded by the curves x=p(y) (on the left) and x=q(y) (on the right) for y

between cand dthen RBf(x, y)dA is also given by RBf(x, y)dA =Rd

cRq(y)

p(y)f(x, y)dx dy.

Example. Let f(x, y) = xy +y2and let Bbe the region between the curves y=x

and y=x2. Compute RBf(x, y)dA.

Solution 1.

The curves y=xand y=x2intersect at x= 0 and x= 1. For xbetween 0 and 1,

x2≤x. Therefore Bis the region between y=x2(below) and y=x(above) for xbetween

0 and 1. Therefore

ZB

f(x, y)dA =Z1

0Z2

x

(xy +y2)dy dx =Z1

0xy2

2+y3

3y=x

y=x2

dx

=Z1

0x3

2+x3

3−x5

2−x6

3dx =Z1

05x3

6−x5

2−x6

3dx

=5x4

24 −x6

12 −x7

211

0

=5

24 −1

12 −1

21 =13

168

Solution 2.

Solving for the curves for xin terms of ygives x=yand x=√yrespectively. The

curves x=yand x=√yintersect at y= 0 and y= 1. For ybetween 0 and 1, y≤√y.

Therefore

ZB

f(x, y)dA =Z1

0Z√y

y

(xy +y2)dx dy =Z1

0x2y

2+xy2x=√y

x=y

dy

=Z1

0y2

2+y5/2−y3

2−y3dy =Z1

0y2

2+y5/2−3

2y3dy

=y3

6−2y7/2

7−3

6y41

0

=1

6−2

7−3

8=13

168

as before. Note that the fact that the outer limits came out to be 0 and 1 after reversing

the order, just as they were before, is a coincidence: the curves meet at the points (0,0)

and (1,1) which happen to have the same xand ycoordinates.

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