Double Integrals — Question 1

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Question 1

Let R=[0,2]×[0,1],f(x,y)=6−x2−y.R=[0,2]\times[0,1],\qquad f(x,y)=6-x^2-y. Partition RR into two equal columns and two equal rows, and use the midpoint of each subrectangle as its sample point.

Tasks

  1. List the four sample points and the common subrectangle area.

  2. Compute the midpoint Riemann sum.

  3. Interpret the sum geometrically and check it against bounds from the minimum and maximum of ff on RR.

Original worksheet page 1: question and worked solution for 4-1-001
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Question 1 – Solution

Strategy. Translate the partition into four base areas, evaluate the height at each midpoint, and add the resulting prism volumes.

See the diagram in the original worksheet below.

Step 1: Partition data Here Δx=1\Delta x=1, Δy=1/2\Delta y=1/2, and ΔA=ΔxΔy=12.\Delta A=\Delta x\Delta y=\frac 12. The midpoints are (12,14),(32,14),(12,34),(32,34).\left(\frac 12,\frac 14\right),\ \left(\frac 32,\frac 14\right),\ \left(\frac 12,\frac 34\right),\ \left(\frac 32,\frac 34\right).

Step 2: Heights and sum In the same order, the four function values are 112,72,5,3.\frac{11}{2},\quad \frac 72,\quad 5,\quad 3. Therefore S=12(112+72+5+3)=172.S=\frac 12\left(\frac{11}{2}+\frac 72+5+3\right)=\boxed{\frac{17}{2}}.

Step 3: Interpretation and bounds The sum is the total volume of four midpoint prisms. On RR, ff decreases as either xx or yy increases, so 1=f(2,1)≤f(x,y)≤f(0,0)=6.1=f(2,1)\le f(x,y)\le f(0,0)=6. Since area⁡(R)=2\operatorname{area}(R)=2, every Riemann-sum estimate lies between 22 and 1212; indeed 17/217/2 does. This also verifies that every sampled height was positive and on the correct scale.

Original worksheet page 2: question and worked solution for 4-1-001

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