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Calculus IIIChapter 4: Multiple IntegralsSection 4-2

Iterated Integrals — Question 10

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Question 10

Over the square R=[−1,1]×[−1,1]R=[-1,1]\times[-1,1], a solid lies between ztop=8−x2−y2andzbottom=x2+y2.z_{\mathrm{top}}=8-x^2-y^2 \qquad\text{and}\qquad z_{\mathrm{bottom}}=x^2+y^2.

Tasks

  1. Verify that the named top surface stays above the bottom surface on all of RR.

  2. Set up and evaluate an iterated integral for the volume.

  3. Check the answer using symmetry and separate one-variable integrals.

Original worksheet page 1: question and worked solution for 4-2-010
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Question 10 – Solution

Strategy. Integrate vertical thickness over the rectangular base, after confirming that the thickness never changes sign.

See the diagram in the original worksheet below.

Step 1: Thickness The vertical thickness is h(x,y)=8−2x2−2y2.h(x,y)=8-2x^2-2y^2. On RR, x2+y2≤2x^2+y^2\le 2, so h≥8−4=4>0h\ge 8-4=4>0. Thus no top–bottom reversal or absolute value is needed.

Step 2: Iterated integral V=∫−11∫−11(8−2x2−2y2)dydx=∫−11(16−4x2−43)dx=[443x−43x3]−11=803.\begin{align*} V&=\int_{-1}^{1}\int_{-1}^{1}(8-2x^2-2y^2)\,dy\,dx\\ &=\int_{-1}^{1}\left(16-4x^2-\frac 43\right)dx\\ &=\left[\frac{44}{3}x-\frac 43x^3\right]_{-1}^{1} =\boxed{\frac{80}{3}}. \end{align*}

Step 3: Independent check The constant term contributes 8area⁡(R)=328\operatorname{area}(R)=32. Also, ∬Rx2dA=(∫−11x2dx)(2)=43,\iint_Rx^2\,dA=\left(\int_{-1}^{1}x^2dx\right)(2)=\frac 43, and identically ∬Ry2dA=4/3\iint_Ry^2\,dA=4/3. Hence V=32−2(43)−2(43)=32−163=803,V=32-2\left(\frac 43\right)-2\left(\frac 43\right) =32-\frac{16}{3}=\frac{80}{3}, confirming the iterated calculation.

Original worksheet page 2: question and worked solution for 4-2-010

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