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Lecture

# Reversing Integration

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University of Toronto Scarborough

Mathematics

MATA33H3

Eric Moore

Fall

Description

Reversing the Order of Integration
Let z = f(x,y) and let B be a subset of the Xy-plane. The volume of the ﬁgure
R
lying below the graph of f(x,y) and over the set B is denBtedx,y)dA To compute
it requires a double integral. If B is the region between the curves y = g(x) (below) and
R RbR h(x)
y = h(x) (above) for x between a and b, thBn(x,y)dA = a g(x)f(x,y)dydx. Note
that the limits of integration depend only the region B and are independent of the function
z = f(x,y) being integrated. Since the preceding discussion is symmetrical with respect to
x and y, we cab reverse the roles of x and y if convenient. That is, if B is written instead
as the region boundeR by the curves x = p(y) (Rn the left) aR Rx = q(y) (on the right) for y
between c and d then f(x,y)dA is also given by f(x,y)dA = d q(yf(x,y)dxdy.
B B c p(y)
2
Example. Let f(x,y) R xy + y and let B be the region between the curves y = x
and y = x . Compute B f(x,y)dA.
Solution 1.
The curves y = x and y = x intersect at x = 0 and x = 1. For x between 0 and 1,
2 2
x ≤ x. Therefore B is the region between y = x (below) and y = x (above) for x between
0 and 1. Therefore
Z Z Z Z
1 2 1 xy2 y3 y=x
f(x,y)dA = (xy + y )dy dx = + dx
B 0 x 0 2 3 y=x2
Z 1 Z 1
x3 x3 x5 x6 5x3 x5 x6
= + − − dx = − − dx
0 2 3 2 3 0 6 2 3
4 6 71
= 5x − x − x = 5 − 1 − 1 = 13
24 12 21 0 24 12 21 168
Solution 2.
√
Solving for the √urves for x in terms of y gives x = y andy respectivel√. The
curves x = y and x = y intersect at y = 0 and y = 1. For y between 0 and 1, y.≤
Therefore
Z Z 1Z √ y Z 1 x= y
2 x y 2
f(x,y)dA = (xy + y )dxdy = + xy dy
B 0 y 0 2 x=y
Z 1 2 3 Z 1 2
= y + y5/2− y − y3 dy = y + y5/2− 3 y3 dy
2 2 2 2
0 0
y3 2y7/2 3 1 1 2 3 13
= − − y4 = − − =
6 7 6 0 6

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