Area & Average Value: Double Integrals - Calculus 3

1. Use a double integral to find the area enclosed by $r=2\sin\theta$.
   Answer: $\pi$ Set up the polar double integral: integrate r dr dθ with r from 0 to 2 sinθ and θ from 0 to π. This equals 1/2 times the integral from 0 to π of (2 sinθ)^2 dθ, which simplifies to 2 times the integral of $\sin^2$ θ, giving π. The circle has diameter 2, so its radius is 1. The tempting wrong answer is 2π, which comes from letting θ run to 2π and tracing the same circle twice.
2. Use a double integral to find the average value of $f(x,y)=x$ on the region bounded by $y=x^2$ and $y=x$.
   Answer: $\frac{1}{2}$ The region has x from 0 to 1, with y from $x^2$ to x. Its area is the integral from 0 to 1 of (x - $x^2$) dx = 1/6. The integral of f(x,y)=x over the region is the integral from 0 to 1 of x(x - $x^2$) dx = 1/12. Dividing gives (1/12)/(1/6) = 1/2. The tempting wrong answer is 1/12, which is the numerator integral before dividing by the area.
3. The region $D$ is defined by $1 \leq x^2 + y^2 \leq 4$ and $0 \leq y \leq x$. What is the area of region $D$?
   Answer: $\frac{3\pi}{8}$ In polar coordinates, $1 \leq r^2 \leq 4$ means $1 \leq r \leq 2$. The condition $0 \leq y \leq x$ with $y = r\sin\theta$ and $x = r\cos\theta$ gives $0 \leq r\sin\theta \leq r\cos\theta$. For $r > 0$, this means $0 \leq \sin\theta \leq \cos\theta$, which occurs when $0 \leq \theta \leq \frac{\pi}{4}$. The area is $\int_0^{\pi/4} \int_1^2 r \, dr \, d\theta = \int_0^{\pi/4} \left[\frac{r^2}{2}\right]_1^2 d\theta = \int_0^{\pi/4} \frac{3}{2} d\theta = \frac{3}{2} \cdot \frac{\pi}{4} = \frac{3\pi}{8}$. Choice B uses the full semicircle, C uses $\theta$ from 0 to $\pi/2$, and D forgets the factor of 3 from the radial integration.
4. A lamina occupies the region between $y = \sin x$ and $y = 0$ for $0 \leq x \leq \pi$. The density at point $(x,y)$ is $\rho(x,y) = 1 + \cos x$. What is the ratio of the average density to the area of the lamina?
   Answer: $\frac{1}{2}$ When you encounter lamina problems involving density functions, you're working with double integrals to find mass and applying the definition of average density. The key insight is that average density equals total mass divided by area. First, let's find the area of the region. The lamina occupies the area between $y = \sin x$ and $y = 0$ from $x = 0$ to $x = \pi$, which gives us: $\text{Area} = \int_0^{\pi} \sin x \, dx = [-\cos x]_0^{\pi} = -\cos(\pi) + \cos(0) = 2$ Next, we calculate the total mass using the density function $\rho(x,y) = 1 + \cos x$: $\text{Mass} = \int_0^{\pi} \int_0^{\sin x} (1 + \cos x) \, dy \, dx = \int_0^{\pi} (1 + \cos x) \sin x \, dx$ Expanding this: $\int_0^{\pi} \sin x \, dx + \int_0^{\pi} \cos x \sin x \, dx = 2 + 0 = 2$ The second integral equals zero because $\cos x \sin x = \frac{1}{2}\sin(2x)$, and integrating from 0 to $\pi$ gives zero. Therefore, the average density is $\frac{\text{Mass}}{\text{Area}} = \frac{2}{2} = 1$, and the ratio of average density to area is $\frac{1}{2}$. Option A ($\frac{2}{\pi}$) incorrectly uses $\pi$ instead of 2 for the area. Option B ($\frac{\pi}{4}$) likely confuses this with a circular region formula. Option D (1) gives just the average density without dividing by the area. Remember: average density problems require careful setup of your integration limits and distinguishing between mass, area, and their ratio.
5. The region $T$ is defined by $0 \leq x \leq 2$, $0 \leq y \leq 2$, and $x + y \leq 2$. What is the average value of $h(x,y) = x^2 + xy + y^2$ over region $T$?
   Answer: $\frac{4}{3}$ When finding the average value of a function over a region, you need to compute $\frac{1}{\text{Area}(T)} \iint_T h(x,y) \, dA$. The key is correctly identifying the region and setting up the bounds of integration. The region $T$ is bounded by $0 \leq x \leq 2$, $0 \leq y \leq 2$, and $x + y \leq 2$. This forms a triangle with vertices at $(0,0)$, $(2,0)$, and $(0,2)$. The area is $\frac{1}{2} \cdot 2 \cdot 2 = 2$. To evaluate $\iint_T (x^2 + xy + y^2) \, dA$, integrate with bounds $0 \leq x \leq 2$ and $0 \leq y \leq 2-x$: $\int_0^2 \int_0^{2-x} (x^2 + xy + y^2) \, dy \, dx$ The inner integral gives: $x^2y + \frac{xy^2}{2} + \frac{y^3}{3}\Big|_0^{2-x} = x^2(2-x) + \frac{x(2-x)^2}{2} + \frac{(2-x)^3}{3}$ Expanding and integrating over $x$ yields $\frac{8}{3}$. Therefore, the average value is $\frac{8/3}{2} = \frac{4}{3}$. Choice A) $2$ incorrectly assumes the area is $4$ (the full square). Choice C) $\frac{8}{3}$ forgets to divide by the area—this is the integral itself, not the average. Choice D) $\frac{5}{3}$ likely results from computational errors in the integration process. Remember: average value problems require both correct region identification and careful integration. Always divide the integral by the region's area, and sketch the region to verify your bounds.
6. Use a double integral to find the average value of $f(x,y)=x$ on the triangle with vertices $(0,0),(1,0),(0,2)$.
   Answer: $\frac{1}{3}$ Integrate x over the triangle: x runs from 0 to 1 and y runs from 0 to 2 - 2x, giving integral of x(2 - 2x) dx = 1/3. The triangle's area is 1, so average = integral divided by area = 1/3. A common error is using area 1/2, which gives 2/3, but base 1 times height 2 makes area 1.
7. A region $R$ in the first quadrant is bounded by $xy = 1$, $xy = 4$, $y = x$, and $y = 2x$. Using the transformation $u = xy$ and $v = \frac{y}{x}$, what is the area of region $R$?
   Answer: $\frac{3\ln 2}{2}$ When you encounter a problem asking for area with a suggested transformation, you're dealing with change of variables in double integrals. The key insight is that transformations can convert complex regions into simple rectangles, making integration much easier. First, let's see what the transformation $u = xy$ and $v = \frac{y}{x}$ does to our boundaries. The curves $xy = 1$ and $xy = 4$ become $u = 1$ and $u = 4$. The lines $y = x$ and $y = 2x$ become $v = 1$ and $v = 2$ (since $\frac{y}{x} = 1$ and $\frac{y}{x} = 2$). So region $R$ transforms into the rectangle $[1,4] \times [1,2]$ in the $uv$-plane. To find the area, we need the Jacobian of the transformation. From $u = xy$ and $v = \frac{y}{x}$, we can solve for $x$ and $y$: $x = \sqrt{\frac{u}{v}}$ and $y = \sqrt{uv}$. Computing the Jacobian determinant gives us $J = \frac{1}{2v}$. The area is $\iint_R 1 \, dx \, dy = \int_1^4 \int_1^2 \frac{1}{2v} \, dv \, du = \int_1^4 \frac{1}{2}[\ln v]_1^2 \, du = \int_1^4 \frac{\ln 2}{2} \, du = \frac{\ln 2}{2} \cdot 3 = \frac{3\ln 2}{2}$. Answer choice (A) $\frac{\ln 2}{2}$ forgets to multiply by the $u$-interval length of 3. Choice (B) $3\ln 2$ omits the factor of $\frac{1}{2}$ from the Jacobian. Choice (C) $\frac{3\ln 2}{4}$ likely results from incorrectly computing the Jacobian as $\frac{1}{4v}$. Remember: always compute the Jacobian determinant carefully when transforming variables, and don't forget to integrate over the entire transformed region.
8. The average value of $f(x,y) = e^{x+y}$ over the square region $[0,a] \times [0,a]$ is $\frac{(e^a - 1)^2}{a^2}$. What is the area of the level curve $f(x,y) = e^a$ that lies within this square?
   Answer: $\frac{a^2}{2}$ The level curve $e^{x+y} = e^a$ simplifies to $x + y = a$. Within the square $[0,a] \times [0,a]$, this line goes from $(0,a)$ to $(a,0)$, dividing the square into two triangular regions. The region where $f(x,y) = e^{x+y} \leq e^a$ (i.e., $x + y \leq a$) is the triangle with vertices $(0,0)$, $(0,a)$, and $(a,0)$. This triangle has area $\frac{1}{2} \cdot a \cdot a = \frac{a^2}{2}$. Choice B gives the total square area, C would be one-fourth of the square, and D would be one-eighth of the square area.
9. Use a double integral to find the area of the region bounded by $y=\sqrt{x}$ and $y=x^2$.
   Answer: $\frac{1}{3}$ The curves intersect at x=0 and x=1, and on that interval sqrt(x) is the upper curve and $x^2$ is the lower curve. So the area is the integral from 0 to 1 of (sqrt(x) - $x^2$) dx, which equals 2/3 - 1/3 = 1/3. A tempting mistake is integrating only sqrt(x) to get 2/3, forgetting to subtract the area under $x^2$.
10. Use a double integral to find the average value of $f(x,y)=x^2$ on the unit disk.
    Answer: $\frac{1}{4}$ Use polar coordinates on the unit disk: x = r cos(theta), area element r dr dtheta. The average is 1/pi times the integral of $r^2$ $\cos^2$(theta) times r. The $r^3$ integral is 1/4 and $\cos^2$ averages to 1/2, so the double integral is pi/4, and dividing by the disk area pi gives 1/4. The tempting 1/2 comes from dropping the $\cos^2$ average.

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