Calculus 3-The Double Integral Over More General Regions, guaranteed 100% Pass

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The Double Integral over more General Regions

Instead of integrating over a rectangle, suppose the region looks like:

𝑦 = 𝑔2 (𝑥) 𝑦 = 𝑔2 (𝑥) 𝑦 = 𝑔2 (𝑥)


𝐷 𝐷
𝐷
𝑦 = 𝑔1 (𝑥)
𝑦 = 𝑔1 (𝑥) 𝑦 = 𝑔1 (𝑥)


These regions are called 𝒚-simple (i.e. 𝑔1 (𝑥 ) ≤ 𝑦 ≤ 𝑔2 (𝑥)).

How do we calculate ∬𝐷 𝑓 (𝑥, 𝑦) 𝑑𝐴? When 𝐷 was a rectangle we did it with an
iterated integral.
𝑥=𝑏 𝑦=𝑑
∬𝐷 𝑓 (𝑥, 𝑦) 𝑑𝐴 = ∫𝑥=𝑎 [∫𝑦=𝑐 𝑓(𝑥, 𝑦) 𝑑𝑦] 𝑑𝑥
𝑑
𝐷
𝑐
𝑎 𝑏

For a region, 𝐷, bounded by 𝑦 = 𝑔2 (𝑥 ), 𝑦 = 𝑔1 (𝑥 ), 𝑥 = 𝑎, 𝑥 = 𝑏,
let 𝑅 be a rectangle containing 𝐷.

𝑦 = 𝑔2 (𝑥) Define:
𝑅 𝐷 𝐹 (𝑥, 𝑦) = 𝑓(𝑥, 𝑦) if (𝑥, 𝑦) ∈ 𝐷
𝑦 = 𝑔1 (𝑥) = 0 if (𝑥, 𝑦) ∈ 𝑅 but not in 𝐷


Define: ∬𝐷 𝑓 (𝑥, 𝑦) 𝑑𝐴 = ∬𝑅 𝐹 (𝑥, 𝑦) 𝑑𝐴. By Fubini’s Theorem:


𝑏 𝑑 𝑏 𝑔2 (𝑥)
∬ 𝐹 (𝑥, 𝑦) 𝑑𝐴 = ∫ ∫ 𝐹 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 = ∫ ∫ 𝑓(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 .
𝑎 𝑐 𝑎 𝑔1 (𝑥)
𝑅

, 2

𝑏 𝑔2 (𝑥)
Notice that ∫𝑎 ∫𝑔 (𝑥) 1 𝑑𝑦 𝑑𝑥 =area between 𝑦 = 𝑔1 (𝑥 ), 𝑦 = 𝑔2 (𝑥),
1
where 𝑎 ≤ 𝑥 ≤ 𝑏.


Ex. Evaluate ∬𝐷 (𝑥 − 2𝑦) 𝑑𝐴, where 𝐷 is the region bounded by:
𝑦 = 2 − 𝑥 2, 𝑦 = 𝑥 2.
𝑦 = 𝑥2


2 − 𝑥2 = 𝑥2 𝐷
2 = 2𝑥 2
𝑥 = ±1

𝑦 = 2 − 𝑥2
𝑥=1 𝑦=2−𝑥 2
∬(𝑥 − 2𝑦) 𝑑𝐴 = ∫ ∫ (𝑥 − 2𝑦) 𝑑𝑦 𝑑𝑥
𝑥=−1 𝑦=𝑥 2
𝐷

1 𝑦=2−𝑥 2
= ∫−1(𝑥𝑦 − 𝑦 2 )|𝑦=𝑥 2 𝑑𝑥

1
= ∫−1(𝑥 (2 − 𝑥 2 ) − (2 − 𝑥 2 )2 ) − (𝑥 (𝑥 2 ) − (𝑥 2 )2 ) 𝑑𝑥

1
= ∫−1 2𝑥 − 𝑥 3 − (4 − 4𝑥 2 + 𝑥 4 ) − (𝑥 3 − 𝑥 4 ) 𝑑𝑥

1
= ∫−1 2𝑥 − 𝑥 3 − 4 + 4𝑥 2 − 𝑥 4 − 𝑥 3 + 𝑥 4 𝑑𝑥

1
= ∫−1 −2𝑥 3 + 4𝑥 2 + 2𝑥 − 4 𝑑𝑥

1 4
= − 𝑥 4 + 𝑥 3 + 𝑥 2 − 4𝑥 ]−11
2 3
1 4 1 4
= (− (1) + + 1 − 4) − (− (1) − + 1 + 4)
2 3 2 3
7 4 9 4 8 16
= − + − ( − ) = −8 + = − .
2 3 2 3 3 3