Double Integrals in Polar Coordinates -...
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Double Integrals in Polar Coordinates Consider a region R described in polar coordinates as R : θ 1 ≤ θ ≤ θ 2 , r 1 ≤ r ≤ r 2 . y x r 1 r 2 θ 1 θ 2 R We can use polar coordinates to evaluate a double integral of a function f (x, y) over region R as follows. ZZ R f (x, y) dA = Z θ 2 θ 1 Z r 2 r 1 f (r cos θ,r sin θ) r dr dθ. Observe that: x = r cos θ, y = r sin θ, and dA = r dr dθ Example 1. Evaluate the double integral ZZ R (x + y) dA where R is the first quadrant region between two circles of radius 1 and 2. x y 1 2 R Solution. In polar coordinates, the region R is described as R : 0 ≤ θ ≤ π 2 , 1 ≤ r ≤ 2. Therefore, ZZ R (x + y) dA = Z π/2 0 Z 2 1 (r cos θ + r sin θ)r dr dθ = Z π/2 0 Z 2 1 r 2 (cos θ + sin θ) dr dθ = Z π/2 0 r 3 3 2 1 (cos θ + sin θ) dθ Gilles Cazelais. Typeset with L A T E X on June 9, 2014.
Transcript of Double Integrals in Polar Coordinates -...
Double Integrals in Polar Coordinates
Consider a region R described in polar coordinates as
R : θ1 ≤ θ ≤ θ2, r1 ≤ r ≤ r2.y
xr1 r2θ1
θ2
R
We can use polar coordinates to evaluate a double integral of a function f(x, y) over region R asfollows. ∫∫
Rf(x, y) dA =
∫ θ2
θ1
∫ r2
r1
f(r cos θ, r sin θ) r dr dθ.
Observe that:x = r cos θ, y = r sin θ, and dA = r dr dθ
Example 1. Evaluate the double integral∫∫R
(x+ y) dA
where R is the first quadrant region between two circles of radius 1 and 2.
x
y
1 2
R
Solution. In polar coordinates, the region R is described as
R : 0 ≤ θ ≤ π
2, 1 ≤ r ≤ 2.
Therefore, ∫∫R
(x+ y) dA =
∫ π/2
0
∫ 2
1(r cos θ + r sin θ)r dr dθ
=
∫ π/2
0
∫ 2
1r2(cos θ + sin θ) dr dθ
=
∫ π/2
0
r3
3
∣∣∣∣∣2
1
(cos θ + sin θ) dθ
Gilles Cazelais. Typeset with LATEX on June 9, 2014.
=7
3
∫ π/2
0(cos θ + sin θ) dθ
=7
3(sin θ − cos θ)
∣∣∣π/20
=7
3(sinπ/2− cosπ/2)− 7
3(sin 0− cos 0)
=14
3
Example 2. Evaluate the double integral ∫∫Rx2 dA
over the region R described in rectangular coordinates by
R : − 3 ≤ x ≤ 3, 0 ≤ y ≤√
9− x2.
Solution. The region R corresponds to the following half-circle.
x
y
3−3 0
R
In polar coordinates it can be described as
R : 0 ≤ θ ≤ π, 0 ≤ r ≤ 3.
Therefore, ∫∫Rx2 dA =
∫ π
0
∫ 3
0(r cos θ)2 r dr dθ
=
∫ π
0
∫ 3
0r3 cos2 θ dr dθ
=
∫ π
0
r4
4
∣∣∣∣∣3
0
cos2 θ dθ
=81
4
∫ π
0cos2 θ dθ
=81
4
∫ π
0
1 + cos(2θ)
2dθ
=81
8
(θ +
sin(2θ)
2
)∣∣∣∣∣π
0
=81π
8
2
Example 3. Find the volume above the xy-plane and under the surface z = 4− x2 − y2.
Solution. The surface z = 4 − x2 − y2 is a paraboloid. It intersects the xy-plane at the level z = 0which is equivalent to the circle of equation x2 + y2 = 4.
y
z
x
z = 4− x2 − y2
Rx
y
2
R
In polar coordinates, we haveR : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2.
Therefore,
V =
∫∫Rz dA
=
∫∫R
(4− x2 − y2) dA
=
∫ 2π
0
∫ 2
0(4− r2) r dr dθ
=
∫ 2π
0
∫ 2
0(4r − r3) dr dθ
=
∫ 2π
0
(2r2 − r4
4
)∣∣∣∣∣2
0
dθ
=
∫ 2π
04 dθ
= 8π
Example 4. Evaluate the following double integral.∫ √20
∫ √4−x2x
(x2 + y2) dy dx
Solution. We have a double integral over a region R described in rectangular coordinates as
R : 0 ≤ x ≤√
2, x ≤ y ≤√
4− x2.
3
y = x
y
x2
2
0
y =√
4− x2R
√2
Evaluating this double integral is much easier if we switch to polar coordinates. In polar coordinates,the line y = x corresponds to θ = π/4 (note that π/4 = 45◦). Therefore, region R can be describedas
R :π
4≤ θ ≤ π
2, 0 ≤ r ≤ 2.
Therefore, ∫ √20
∫ √4−x2x
(x2 + y2) dy dx =
∫ π/2
π/4
∫ 2
0(r2) r dr dθ
=
∫ π/2
π/4
∫ 2
0r3 dr dθ
=
∫ π/2
π/4
r4
4
∣∣∣∣∣2
0
dθ
=
∫ π/2
π/44 dθ
= 4(π
2− π
4
)= π
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