Calculus 3-The Double Integral Over a Rectangle, guaranteed 100% Pass

1


Double Integrals Over Rectangles

Recall for 1 variable: 𝑦 = 𝑓(𝑥)

𝑦 = 𝑓(𝑥)

𝑏−𝑎
= Δ𝑥
𝑛
𝑏
∫𝑎 𝑓(𝑥) 𝑑𝑥 = lim ∑𝑛𝑖=1 𝑓(𝑥𝑖 ) Δ𝑥
𝑛→∞

where 𝑥𝑖 = 𝑎 + 𝑖Δ𝑥



𝑎 𝑥1 𝑥2 𝑥3 𝑥𝑛 = 𝑏


𝑏
∫𝑎 𝑓(𝑥) 𝑑𝑥 = area under the curve if 𝑓(𝑥 ) ≥ 0.
Fundamental Theorem of Calculus:
𝑏
∫𝑎 𝑓(𝑥) 𝑑𝑥 = 𝐹 (𝑏) − 𝐹(𝑎) if 𝐹 ′ (𝑥 ) = 𝑓(𝑥 ).


For functions of 2 variables we start with a closed rectangle, 𝑅:
𝑅 = [𝑎, 𝑏] × [𝑐, 𝑑] = {(𝑥, 𝑦) ∈ ℝ2 | 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑}
First, let’s assume 𝑓 (𝑥, 𝑦) ≥ 0
Let 𝑆 be the solid that lies above 𝑅 and under the
graph of 𝑧 = 𝑓(𝑥, 𝑦).
𝑆 = {(𝑥, 𝑦, 𝑧)| 0 ≤ 𝑧 ≤ 𝑓(𝑥, 𝑦); (𝑥, 𝑦) ∈ 𝑅}.
We want to find the volume of 𝑆.



𝑆

𝑅

, 2


First we partition 𝑅 into subrectangles. We do this by dividing up the intervals
[𝑎, 𝑏] and [𝑐, 𝑑].

𝑎 = 𝑥0 < 𝑥1 < 𝑥2 < ⋯ < 𝑥𝑖 < ⋯ < 𝑥𝑛 = 𝑏

𝑐 = 𝑦0 < 𝑦1 < 𝑦2 < ⋯ < 𝑦𝑗 < ⋯ < 𝑦𝑚 = 𝑑

𝑅
𝑑 = 𝑦𝑚




𝑦𝑗
(𝑥𝑖∗ , 𝑦𝑗∗ )
𝑦1

𝑐 = 𝑦0
𝑅𝑖𝑗

𝑎 = 𝑥0 𝑥1 𝑥𝑖 𝑏 = 𝑥𝑛


𝑅𝑖𝑗 = [𝑥𝑖−1 , 𝑥𝑖 ] × [𝑦𝑗−1 , 𝑦𝑗 ] = {(𝑥, 𝑦)| 𝑥𝑖−1 ≤ 𝑥 ≤ 𝑥𝑖 , 𝑦𝑗−1 ≤ 𝑦 ≤ 𝑦𝑗 }

Area of 𝑅𝑖𝑗 is Δ𝐴𝑖𝑗 = (Δ𝑥𝑖 )(Δ𝑦𝑗 )

Choose any point in each rectangle 𝑅𝑖𝑗 and call it (𝑥𝑖∗ , 𝑦𝑗∗ ).