Calculus- Early Transcendentals, 2021a

14.2. Double Integrals in Polar Coordinates 495
and θ, we use small pieces of ring-shaped areas, as shown in Figure 14.7. Each small region is roughly
rectangular, except that two sides are segments of a circle and the other two sides are not quite parallel.
Near a point (r,θ), the length of either circular arc is about rΔθ and the length of each straight side is
simply Δr. When Δr and Δθ are very small, the region is nearly a rectangle with area rΔrΔθ, andthe
volume under the surface is approximately
∑∑ f (r i ,θ j )r i ΔrΔθ.
In the limit, this turns into a double integral
∫ θ1
∫ r1
θ 0 r 0
f (r,θ)rdrdθ.
.
.
.
.
rΔθ
Δr
.
Figure 14.7: A polar coordinates “grid”.
Example 14.4: Volume of One-Eighth of a Sphere
Find the volume under z = √ 4 − r 2 above the quarter circle bounded by the two axes and the circle
x 2 + y 2 = 4 in the first quadrant.
Solution. In terms of r and θ, this region is described by the restrictions 0 ≤ r ≤ 2and0≤ θ ≤ π/2, so
we have
∫ π/2 ∫ 2 √
∫ π/2
4 − r 2 rdrdθ = − 1 ∣ ∣∣∣
3 (4 − r2 ) 3/2 2
dθ
0
0
=
0
∫ π/2
0
= 4π 3 .
8
3 dθ
The surface is a portion of the sphere of radius 2 centered at the origin, in fact exactly one-eighth of the
sphere. We know the formula for volume of a sphere is (4/3)πr 3 , so the volume we have computed is
(1/8)(4/3)π2 3 =(4/3)π, in agreement with our answer.
♣
0