Iterated Integrals — Question 10

Question 10

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

Tasks

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

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

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

Question 10 – Solution

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

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

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

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