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Calculus IIIChapter 4: Multiple IntegralsSection 4-1

Double Integrals — Question 7

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Question 7

The plane z=6+x−2yz=6+x-2y lies above the rectangle R=[0,2]×[0,1].R=[0,2]\times[0,1].

Tasks

  1. Pair each point of RR with its reflection through the center (1,1/2)(1,1/2) and compare their heights.

  2. Use the pairing to find the volume under the plane without iterated integration.

  3. Verify the result using the four corner heights.

Original worksheet page 1: question and worked solution for 4-1-007
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Question 7 – Solution

Strategy. An affine function has complementary values at points reflected through a rectangle’s center, so every reflected pair has the same average height.

See the diagram in the original worksheet below.

Step 1: Reflected heights The reflection of P=(x,y)P=(x,y) through C=(1,1/2)C=(1,1/2) is P*=(2−x,1−y).P^*=(2-x,1-y). For f(x,y)=6+x−2yf(x,y)=6+x-2y, f(P)+f(P*)=(6+x−2y)+(6+(2−x)−2(1−y))=12.\begin{align*} f(P)+f(P^*) &=(6+x-2y)+\bigl(6+(2-x)-2(1-y)\bigr)\\ &=12. \end{align*} Thus every pair has average height 6=f(C)6=f(C).

Step 2: Volume The base area is 22, and the plane’s minimum corner height is 4>04>0, so the double integral is geometric volume: V=∬R(6+x−2y)dA=6(2)=12.\boxed{V=\iint_R(6+x-2y)\,dA=6(2)=12}.

Step 3: Corner check The corner heights are f(0,0)=6,f(2,0)=8,f(0,1)=4,f(2,1)=6.f(0,0)=6,\quad f(2,0)=8,\quad f(0,1)=4,\quad f(2,1)=6. Their average is (6+8+4+6)/4=6(6+8+4+6)/4=6, agreeing with the center and pairing calculation. Multiplication by base area again gives 1212.

Original worksheet page 2: question and worked solution for 4-1-007

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