Double Integrals — Question 6

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Question 6

On the centrally symmetric rectangle R=[−2,2]×[−1,1],R=[-2,2]\times[-1,1], consider f(x,y)=5+x3+xy2+2yf(x,y)=5+x^3+xy^2+2y.

Tasks

  1. Show that the nonconstant part cancels under central reflection.

  2. Evaluate ∬RfdA\iint_Rf\,dA and find the average value of ff.

  3. Explain precisely which symmetry of the region makes the argument valid.

Original worksheet page 1: question and worked solution for 4-1-006
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Question 6 – Solution

Strategy. Pair each point (x,y)(x,y) with its central reflection (−x,−y)(-x,-y) and average the two function values.

Step 1: Pairing Let h(x,y)=x3+xy2+2yh(x,y)=x^3+xy^2+2y. Then h(−x,−y)=−x3−xy2−2y=−h(x,y).h(-x,-y)=-x^3-xy^2-2y=-h(x,y). Thus the contributions of hh at every reflected pair cancel. Equivalently, f(x,y)+f(−x,−y)=10.f(x,y)+f(-x,-y)=10.

Step 2: Integral and average Central reflection maps RR onto itself without changing area. Therefore the paired average of ff is 55 everywhere, and ∬RfdA=5area⁡(R)=5(4)(2)=40.\iint_Rf\,dA=5\operatorname{area}(R)=5(4)(2)=\boxed{40}. Dividing by the area 88 gives favg=5.\boxed{f_{\mathrm{avg}}=5}.

Step 3: Symmetry check The required property is (x,y)∈R⇔(−x,−y)∈R.(x,y)\in R\quad\Longleftrightarrow\quad(-x,-y)\in R. Symmetry in only one coordinate would not automatically cancel the entire hh: xy2xy^2 is odd in xx, while 2y2y is odd in yy. Central symmetry reverses both coordinates at once and cancels all three terms.

Original worksheet page 2: question and worked solution for 4-1-006

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