Area & Average Value: Double Integrals - Multivariable Calculus

1. Consider the region bounded by $x = 0$, $y = 0$, and $x + 2y = 6$ in the first quadrant. If the average value of $f(x,y) = x^2 + y^2$ over this region is $k$, then the average value of $g(x,y) = 2x^2 + 2y^2$ over the same region is:
   Answer: $2k$ This problem tests understanding of the linearity property of average values. The average value of a function $f$ over a region $R$ is $\bar{f} = \frac{1}{\text{Area}(R)} \iint_R f(x,y) \, dA$. For the function $g(x,y) = 2x^2 + 2y^2 = 2f(x,y)$, we have: $\bar{g} = \frac{1}{\text{Area}(R)} \iint_R g(x,y) \, dA = \frac{1}{\text{Area}(R)} \iint_R 2f(x,y) \, dA = \frac{2}{\text{Area}(R)} \iint_R f(x,y) \, dA = 2 \cdot \frac{1}{\text{Area}(R)} \iint_R f(x,y) \, dA = 2\bar{f} = 2k$. This follows from the linearity property of integrals: $\int cf(x) dx = c\int f(x) dx$. Choice B ($4k$) might be chosen by students who incorrectly think that since we're squaring terms, the average gets squared. Choice C ($k + 2$) reflects a misunderstanding of how constants affect averages. Choice D ($k^2$) comes from incorrectly squaring the entire average value.
2. The area of the region enclosed by the polar curve $r = 2 + \cos(3\theta)$ can be expressed as a double integral. Which of the following correctly represents this area?
   Answer: $\int_0^{2\pi} \int_0^{2+\cos(3\theta)} r \, dr \, d\theta$ In polar coordinates, the area element is $dA = r \, dr \, d\theta$. For a region bounded by $r = f(\theta)$, the area is $\iint_R dA = \int_{\theta_1}^{\theta_2} \int_0^{f(\theta)} r \, dr \, d\theta$. The curve $r = 2 + \cos(3\theta)$ has period $\frac{2\pi}{3}$ but traces the complete curve once as $\theta$ goes from $0$ to $2\pi$, creating a three-petaled rose-like shape. Since $2 + \cos(3\theta) \geq 1 > 0$ for all $\theta$, the curve doesn't pass through the origin and we integrate from $0$ to $2\pi$ to capture the entire region. Choice B uses wrong limits ($0$ to $\pi$ would miss part of the curve). Choice C is missing the $r$ factor needed for the polar area element. Choice D incorrectly includes a factor of $\frac{1}{2}$, which would be used for areas swept by radial lines, not enclosed regions.
3. A region $R$ is bounded by the curves $y = x^2$ and $y = 2x$. If the average value of the function $f(x,y) = xy$ over region $R$ is $\frac{8}{5}$, what is the area of region $R$?
   Answer: $\frac{4}{3}$ First, find the intersection points: $x^2 = 2x$ gives $x = 0$ and $x = 2$. The region is bounded by $0 \leq x \leq 2$ and $x^2 \leq y \leq 2x$. The average value formula is $\bar{f} = \frac{1}{A} \iint_R f(x,y) \, dA$, so $A = \frac{\iint_R xy \, dA}{\bar{f}}$. Computing $\iint_R xy \, dA = \int_0^2 \int_{x^2}^{2x} xy \, dy \, dx = \int_0^2 x \cdot \frac{(2x)^2 - (x^2)^2}{2} \, dx = \int_0^2 x \cdot \frac{4x^2 - x^4}{2} \, dx = \int_0^2 \frac{4x^3 - x^5}{2} \, dx = \frac{1}{2}[x^4 - \frac{x^6}{6}]_0^2 = \frac{1}{2}(16 - \frac{64}{6}) = \frac{16}{15}$. Therefore, $A = \frac{16/15}{8/5} = \frac{16}{15} \cdot \frac{5}{8} = \frac{4}{3}$.
4. Consider the region $R$ defined by $\{(x,y): 1 \leq x^2 + y^2 \leq 4, x \geq 0, y \geq 0\}$. The average value of $f(x,y) = x^2 + y^2$ over this region is:
   Answer: $\frac{5}{2}$ The region is a quarter annulus in the first quadrant. Using polar coordinates: $1 \leq r \leq 2$, $0 \leq \theta \leq \frac{\pi}{2}$, and $f(x,y) = r^2$. The area is $A = \int_0^{\pi/2} \int_1^2 r \, dr \, d\theta = \frac{\pi}{2} \cdot \left[\frac{r^2}{2}\right]_1^2 = \frac{\pi}{2} \cdot \frac{3}{2} = \frac{3\pi}{4}$. The integral is $\iint_R r^2 \cdot r \, dr \, d\theta = \int_0^{\pi/2} \int_1^2 r^3 \, dr \, d\theta = \frac{\pi}{2} \cdot \left[\frac{r^4}{4}\right]_1^2 = \frac{\pi}{2} \cdot \frac{15}{4} = \frac{15\pi}{8}$. The average value is $\frac{15\pi/8}{3\pi/4} = \frac{15\pi}{8} \cdot \frac{4}{3\pi} = \frac{5}{2}$.
5. The region $D$ is defined by $\{(x,y): x^2 + y^2 \leq 9, x \geq 0\}$. If $\iint_D f(x,y) \, dA = 18\pi$ and the area of $D$ is $\frac{9\pi}{2}$, what is the average value of $f(x,y)$ over region $D$?
   Answer: $4$ The average value of a function $f(x,y)$ over a region $D$ is given by $\bar{f} = \frac{1}{\text{Area}(D)} \iint_D f(x,y) \, dA$. We are given that $\iint_D f(x,y) \, dA = 18\pi$ and the area of $D$ is $\frac{9\pi}{2}$. Therefore: $\bar{f} = \frac{18\pi}{9\pi/2} = \frac{18\pi \cdot 2}{9\pi} = \frac{36\pi}{9\pi} = 4$. Note that region $D$ is a semicircle of radius 3 in the right half-plane, so its area is indeed $\frac{1}{2} \cdot \pi \cdot 3^2 = \frac{9\pi}{2}$, confirming the given information.
6. The area of the region $R = \{(x,y): 0 \leq x \leq 2, x^2 \leq y \leq 2x\}$ can be computed as $\int_0^2 \int_{x^2}^{2x} dy \, dx$. If we want to express this same area using the order $dx \, dy$, which of the following integrals is correct?
   Answer: $\int_0^2 \int_{y/2}^{\sqrt{y}} dx \, dy + \int_2^4 \int_{0}^{y/2} dx \, dy$ To change the order of integration, we need to determine the range of $y$ values and express the $x$ boundaries as functions of $y$. The region is bounded by $y = x^2$ and $y = 2x$ for $0 \leq x \leq 2$. These curves intersect when $x^2 = 2x$, giving $x = 0$ and $x = 2$. The $y$ values range from 0 to 4 (since $y = 2x$ reaches $y = 4$ when $x = 2$). For $0 \leq y \leq 2$: From $y = x^2$, we get $x = \sqrt{y}$ (right boundary). From $y = 2x$, we get $x = y/2$ (left boundary). So $y/2 \leq x \leq \sqrt{y}$. For $2 \leq y \leq 4$: The curve $y = x^2$ no longer bounds the region (since $x^2 > 2$ only for $x > \sqrt{2} \approx 1.41$, but the region ends at $x = 2$). The left boundary is $x = 0$ and the right boundary is $x = y/2$ (from $y = 2x$). So $0 \leq x \leq y/2$. Therefore: $\text{Area} = \int_0^2 \int_{y/2}^{\sqrt{y}} dx \, dy + \int_2^4 \int_0^{y/2} dx \, dy$. Choice A uses wrong limits and doesn't split the integral. Choice B has the boundaries reversed in the first integral. Choice D also has reversed boundaries in the first integral.
7. A lamina occupies the triangular region with vertices at $(0,0)$, $(4,0)$, and $(0,3)$. If the density function is $\rho(x,y) = x + y$, what is the average density of the lamina?
   Answer: $\frac{7}{3}$ The triangular region has vertices at $(0,0)$, $(4,0)$, and $(0,3)$. The hypotenuse has equation $\frac{x}{4} + \frac{y}{3} = 1$, or $y = 3 - \frac{3x}{4}$. The region is described by $0 \leq x \leq 4$ and $0 \leq y \leq 3 - \frac{3x}{4}$. The average density is $\bar{\rho} = \frac{\iint_R \rho(x,y) \, dA}{\text{Area}}$. The area of the triangle is $\frac{1}{2} \cdot 4 \cdot 3 = 6$. Now compute $\iint_R (x+y) \, dA = \int_0^4 \int_0^{3-3x/4} (x+y) \, dy \, dx = \int_0^4 \left[xy + \frac{y^2}{2}\right]_0^{3-3x/4} dx = \int_0^4 \left[x(3-\frac{3x}{4}) + \frac{(3-3x/4)^2}{2}\right] dx = \int_0^4 \left[3x - \frac{3x^2}{4} + \frac{(3-3x/4)^2}{2}\right] dx$. Expanding $(3-3x/4)^2 = 9 - \frac{18x}{4} + \frac{9x^2}{16} = 9 - \frac{9x}{2} + \frac{9x^2}{16}$, so $\frac{(3-3x/4)^2}{2} = \frac{9}{2} - \frac{9x}{4} + \frac{9x^2}{32}$. The integrand becomes: $3x - \frac{3x^2}{4} + \frac{9}{2} - \frac{9x}{4} + \frac{9x^2}{32} = \frac{9}{2} + 3x - \frac{9x}{4} - \frac{3x^2}{4} + \frac{9x^2}{32} = \frac{9}{2} + \frac{3x}{4} - \frac{24x^2}{32} + \frac{9x^2}{32} = \frac{9}{2} + \frac{3x}{4} - \frac{15x^2}{32}$. Integrating: $\int_0^4 \left[\frac{9}{2} + \frac{3x}{4} - \frac{15x^2}{32}\right] dx = \left[\frac{9x}{2} + \frac{3x^2}{8} - \frac{15x^3}{96}\right]_0^4 = 18 + 6 - \frac{15 \cdot 64}{96} = 24 - 10 = 14$. Therefore, $\bar{\rho} = \frac{14}{6} = \frac{7}{3}$.

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