Double Integrals — Question 4

Question 4 – Solution

Strategy. Split the base into the two constant-height bands and distinguish algebraic volume from total unsigned volume.

Step 1: Signed volume The lower band has area $4(1)=4$ and height $3$, so it contributes $12$. The upper band has area $4(2)=8$ and height $-2$, so it contributes $-16$. Hence $$\boxed{\iint_R f\,dA=12-16=-4}.$$

Step 2: Unsigned volume Absolute value changes the second height from $-2$ to $2$. Therefore $$\boxed{\iint_R|f|\,dA=12+16=28}.$$ The first integral records volume below the $xy$-plane negatively; the second records all geometric volume positively.

Step 3: Average value Since $\operatorname{area}(R)=4\cdot 3=12$, $$\boxed{f_{\mathrm{avg}}=\frac 1{12}\iint_R f\,dA=-\frac 13}.$$ As a check, $-1/3$ lies between the only two function values, $-2$ and $3$, and the wider negative band explains its negative sign.