Multivariate Calculus - Bruce E. Shapiro

138 LECTURE 17. DOUBLE INTEGRALS OVER GENERAL REGIONS
Solution. The equation of a circle is
Solving for y,
x 2 + y 2 = a 2
y = ± √ a 2 − x 2
The set S can be described as follows: As x increases from x = 0 to x = a, y at x
increases from y = 0 to y = √ a 2 − x 2 . Therefore
s f(x, y)dA = ∫ a
0
∫ √ a 2 −x 2
0
f(x, y)dydx
Example 17.3 Find a formula for the area of a circle of radius a using double
integrals.
Solution. We can place the center of the circle at the origin, so that its equation is
given by x 2 + y 2 = a 2 , as in the previous example. The circle can be thought of as
the set
{(x, y) : −a ≤ x ≤ a, − √ a 2 − x 2 ≤ y ≤ √ a 2 − x 2 }
so that
A =
=
=
= 2
R dxdy = ∫ a
∫ a
−a
∫ a
(
−a
∫ a
−a
−a
√ )
y| a 2 −x 2
− √ a 2 −x 2
∫ √ a 2 −x 2
− √ a 2 −x 2 dydx
dx
(√
a 2 − x 2 − − √ a 2 − x 2 )
dx
√
a 2 − x 2 dx
By symmetry, since the integrand is even, we also have
∫ a √
A = 4 a 2 − x 2 dx
According to integral formula (54) on the inside book jacket
∫ √a 2 − x 2 du = x 2
√
a 2 − x 2 + x2
2 sin−1 x a
Therefore
A = 4
= 4
0
∫ a √
( x √
a 2 − x 2 dx = a
0
2 2 − x 2 + x2 x )∣ ∣∣∣
a
2 sin−1 a
0
[( a a
2√ 2 − a 2 + a2 a ) ( 0
2 sin−1 − a
a 2√ 2 − 0 2 + 02 0 )]
2 sin−1 a
= 4[0 + a2
2 sin−1 (1) − 0 − 0] = 4(a 2 /2)(π/2)
= πa 2
Revised December 6, 2006. Math 250, Fall 2006