Double Integrals in Polar - Multivariable Calculus

Which of the following double integrals represents the area of the region in the first quadrant that lies inside the circle $r=4\sin\theta$ but outside the circle $r=2$ ?
Answer: $\int_{\pi/6}^{\pi/2} \int_2^{4\sin\theta} r \, dr \, d\theta$ When setting up double integrals in polar coordinates to find areas between curves, you need to carefully determine both the angular limits and the radial limits by analyzing where the curves intersect and which curve forms the outer boundary. First, let's understand our region. The circle $r = 4\sin\theta$ can be rewritten in Cartesian form as $x^2 + (y-2)^2 = 4$ , which is a circle centered at $(0,2)$ with radius 2. The circle $r = 2$ is centered at the origin with radius 2. These circles intersect where $2 = 4\sin\theta$ , giving us $\sin\theta = \frac{1}{2}$ , so $\theta = \frac{\pi}{6}$ . For the angular limits, our region exists from $\theta = \frac{\pi}{6}$ (the intersection point) to $\theta = \frac{\pi}{2}$ (where the upper circle reaches its maximum). For radial limits, we integrate from the inner circle $r = 2$ to the outer circle $r = 4\sin\theta$ . The area element in polar coordinates is $r \, dr \, d\theta$ . Choice A uses incorrect angular limits starting from 0 instead of $\frac{\pi}{6}$ . Choice B also starts from 0, missing the intersection point entirely. Choice C has the radial limits reversed, integrating from the outer curve to the inner curve, which would give a negative result. Choice D correctly captures the region with $\theta$ from $\frac{\pi}{6}$ to $\frac{\pi}{2}$ and $r$ from 2 to $4\sin\theta$ . Study tip: Always find intersection points first to determine your integration limits, and remember that in polar coordinates, you typically integrate from inner to outer curve for the radial component.
The integral $\int_0^a \int_0^{\sqrt{a^2-x^2}} f(x^2+y^2) \, dy \, dx$ is converted to polar coordinates. If the resulting integral is $\int_0^{\alpha} \int_0^a f(r^2) r \, dr \, d\theta$ , what is $\alpha$ ?
Answer: $\frac{\pi}{2}$ The original region is described by $0 \leq x \leq a$ and $0 \leq y \leq \sqrt{a^2-x^2}$ . The constraint $y \leq \sqrt{a^2-x^2}$ means $y^2 \leq a^2-x^2$ , so $x^2+y^2 \leq a^2$ . Combined with $x \geq 0$ and $y \geq 0$ , this describes a quarter-circle of radius $a$ in the first quadrant. In polar coordinates, this corresponds to $0 \leq r \leq a$ and $0 \leq \theta \leq \frac{\pi}{2}$ . Therefore, $\alpha = \frac{\pi}{2}$ . Choice A would represent only an eighth of the circle. Choice C would be a semicircle. Choice D would be the full circle.
Let $R$ be the region in the $xy$ -plane bounded by the limaçon $r=2+\sin\theta$ . Evaluate the integral $\iint_R x \, dA$ .
Answer: $0$ The region of integration $R$ is bounded by $r=2+\sin\theta$ . Since $\sin(\pi - \theta) = \sin\theta$ , the region is symmetric with respect to the $y$ -axis. The integrand is $f(x,y) = x$ . This is an odd function with respect to $x$ , meaning $f(-x, y) = -f(x, y)$ . When integrating an odd function with respect to $x$ over a region symmetric with respect to the $y$ -axis, the result is 0. Explicitly, the integral is: $\iint_R x \, dA = \int_0^{2\pi} \int_0^{2+\sin\theta} (r\cos\theta) \cdot r \, dr \, d\theta = \int_0^{2\pi} \cos\theta \left[ \frac{r^3}{3} \right]_0^{2+\sin\theta} d\theta = \frac{1}{3} \int_0^{2\pi} (2+\sin\theta)^3 \cos\theta \, d\theta$ Let $u = 2+\sin\theta$ , so $du = \cos\theta \, d\theta$ . When $\theta=0$ , $u=2$ . When $\theta=2\pi$ , $u=2$ . The integral becomes $\frac{1}{3} \int_2^2 u^3 \, du = 0$ .
Consider the double integral $\iint_R (x^2 + y^2) \, dA$ where $R$ is the region bounded by the circles $x^2 + y^2 = 1$ and $x^2 + y^2 = 9$ , and the rays $\theta = \frac{\pi}{6}$ and $\theta = \frac{\pi}{3}$ . After converting to polar coordinates, which of the following represents the correct setup?
Answer: $\int_{\pi/6}^{\pi/3} \int_1^3 r^2 \cdot r \, dr \, d\theta$ In polar coordinates, $x^2 + y^2 = r^2$ and $dA = r \, dr \, d\theta$ . The region has $1 \leq r \leq 3$ (not 9, since $r = \sqrt{x^2 + y^2}$ ) and $\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}$ . The integrand becomes $r^2 \cdot r = r^3$ . Choice B incorrectly uses $r = 9$ instead of $r = 3$ . Choice C has the correct limits but wrong order of integration for the standard setup. Choice D doesn't simplify the integrand and omits the Jacobian $r$ .
Find the volume of the solid lying under the surface $z = \sqrt{x^2+y^2}$ and above the region $R$ in the $xy$ -plane enclosed by one leaf of the rose $r=4\sin(2\theta)$ in the first quadrant.
Answer: $\frac{128}{9}$ The surface is $z=r$ . The region $R$ corresponding to the leaf in the first quadrant is traced for $0 \le \theta \le \pi/2$ . The volume is given by the double integral of $z$ over $R$ . The area element is $dA = r \, dr \, d\theta$ . $V = \iint_R z \, dA = \int_0^{\pi/2} \int_0^{4\sin(2\theta)} r \cdot r \, dr \, d\theta = \int_0^{\pi/2} \int_0^{4\sin(2\theta)} r^2 \, dr \, d\theta$ The inner integral is $\int_0^{4\sin(2\theta)} r^2 \, dr = [\frac{r^3}{3}]_0^{4\sin(2\theta)} = \frac{64}{3}\sin^3(2\theta)$ . The outer integral is: $V = \frac{64}{3} \int_0^{\pi/2} \sin^3(2\theta) \, d\theta$ Let $u=2\theta$ , $du=2d\theta$ . The integral becomes $\frac{64}{3} \int_0^{\pi} \sin^3(u) \frac{du}{2} = \frac{32}{3} \int_0^{\pi} (1-\cos^2u)\sin u \, du$ . Let $v=\cos u, dv=-\sin u \, du$ . $V = \frac{32}{3} \int_{1}^{-1} (1-v^2) (-dv) = \frac{32}{3} \int_{-1}^{1} (1-v^2) \, dv = \frac{32}{3} [v-\frac{v^3}{3}]_{-1}^1 = \frac{32}{3} [(1-\frac{1}{3}) - (-1+\frac{1}{3})] = \frac{32}{3} [\frac{2}{3} - (-\frac{2}{3})] = \frac{32}{3} \cdot \frac{4}{3} = \frac{128}{9}$
Evaluate the definite integral $\int_0^2 \int_0^{\sqrt{4-x^2}} (x^2+y^2)^{3/2} \, dy \, dx$ .
Answer: $\frac{16\pi}{5}$ The region of integration is described by $0 \le x \le 2$ and $0 \le y \le \sqrt{4-x^2}$ , which is a quarter-circle of radius 2 in the first quadrant. In polar coordinates, this region is $0 \le r \le 2$ and $0 \le \theta \le \pi/2$ . The integrand $(x^2+y^2)^{3/2}$ becomes $(r^2)^{3/2} = r^3$ . The area element $dy \, dx$ becomes $r \, dr \, d\theta$ . The integral is transformed to: $\int_0^{\pi/2} \int_0^2 (r^3) \cdot r \, dr \, d\theta = \int_0^{\pi/2} \int_0^2 r^4 \, dr \, d\theta$ Evaluating the inner integral: $\int_0^2 r^4 \, dr = \left[ \frac{r^5}{5} \right]_0^2 = \frac{32}{5}$ Evaluating the outer integral: $\int_0^{\pi/2} \frac{32}{5} \, d\theta = \frac{32}{5} [\theta]_0^{\pi/2} = \frac{32}{5} \cdot \frac{\pi}{2} = \frac{16\pi}{5}$
Evaluate the improper integral $\iint_R \frac{1}{(1+x^2+y^2)^2} \, dA$ where $R$ is the first quadrant of the $xy$ -plane.
Answer: $\frac{\pi}{4}$ In polar coordinates, the first quadrant is described by $0 \le r < \infty$ and $0 \le \theta \le \pi/2$ . The integrand becomes $\frac{1}{(1+r^2)^2}$ , and $dA = r \, dr \, d\theta$ . The integral is: $I = \int_0^{\pi/2} \int_0^{\infty} \frac{1}{(1+r^2)^2} r \, dr \, d\theta$ First, evaluate the inner improper integral: $\int_0^{\infty} \frac{r}{(1+r^2)^2} \, dr$ . Use the substitution $u = 1+r^2$ , so $du = 2r \, dr$ . As $r \to 0$ , $u \to 1$ . As $r \to \infty$ , $u \to \infty$ . $\int_1^{\infty} \frac{1}{u^2} \frac{du}{2} = \frac{1}{2} \left[ -\frac{1}{u} \right]_1^{\infty} = \frac{1}{2} \left( \lim_{b \to \infty} -\frac{1}{b} - (-\frac{1}{1}) \right) = \frac{1}{2}(0+1) = \frac{1}{2}$ Now, evaluate the outer integral: $I = \int_0^{\pi/2} \frac{1}{2} \, d\theta = \frac{1}{2} [\theta]_0^{\pi/2} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$
Consider evaluating $\int_0^{1} \int_0^{\sqrt{1-y^2}} e^{-(x^2+y^2)} \, dx \, dy$ using polar coordinates. After the conversion and evaluation of the inner integral, what expression must be integrated with respect to $\theta$ ?
Answer: $\frac{1}{2}(1-e^{-1})$ When you encounter a double integral with $x^2 + y^2$ in the integrand and a region that suggests circular boundaries, converting to polar coordinates is usually the most efficient approach. The region of integration here is bounded by $x = \sqrt{1-y^2}$ , which describes a quarter-circle of radius 1 in the first quadrant. In polar coordinates, we have $x = r\cos\theta$ , $y = r\sin\theta$ , and $dx\,dy = r\,dr\,d\theta$ . The region becomes $0 \leq r \leq 1$ and $0 \leq \theta \leq \frac{\pi}{2}$ . The integral transforms to: $\int_0^{\pi/2} \int_0^{1} e^{-r^2} \cdot r \, dr \, d\theta$ Evaluating the inner integral with respect to $r$ : Let $u = -r^2$ , so $du = -2r\,dr$ , giving us $r\,dr = -\frac{1}{2}du$ . When $r = 0$ , $u = 0$ ; when $r = 1$ , $u = -1$ . This yields: $\int_0^{1} re^{-r^2} \, dr = -\frac{1}{2}\int_0^{-1} e^u \, du = -\frac{1}{2}[e^u]_0^{-1} = -\frac{1}{2}(e^{-1} - 1) = \frac{1}{2}(1 - e^{-1})$ This expression is independent of $\theta$ , confirming answer C. Answer A incorrectly includes $\cos\theta$ , D incorrectly includes $\sin\theta$ , and B has the wrong sign ( $e^{-1} - 1$ instead of $1 - e^{-1}$ ). Strategy tip: When the inner integral in polar coordinates yields a constant (no $\theta$ dependence), that constant gets integrated over the $\theta$ range. Always check whether your result depends on the angular variable.
Find the area of the region that lies inside the circle $r=3\cos\theta$ and outside the cardioid $r=1+\cos\theta$ .
Answer: $\pi$ First, find the intersection points of the two curves: $3\cos\theta = 1+\cos\theta \implies 2\cos\theta = 1 \implies \cos\theta = 1/2$ . This occurs at $\theta = -\pi/3$ and $\theta = \pi/3$ . The region is symmetric about the polar axis, so we can integrate from $0$ to $\pi/3$ and multiply by 2. The area is given by $A = \frac{1}{2} \int_{\alpha}^{\beta} (r_{outer}^2 - r_{inner}^2) \, d\theta$ . In the interval $[-\pi/3, \pi/3]$ , $3\cos\theta \ge 1+\cos\theta$ . Thus, $r_{outer} = 3\cos\theta$ and $r_{inner} = 1+\cos\theta$ . $A = 2 \cdot \frac{1}{2} \int_0^{\pi/3} [(3\cos\theta)^2 - (1+\cos\theta)^2] \, d\theta = \int_0^{\pi/3} [9\cos^2\theta - (1+2\cos\theta+\cos^2\theta)] \, d\theta$ $= \int_0^{\pi/3} (8\cos^2\theta - 2\cos\theta - 1) \, d\theta$ Using $\cos^2\theta = \frac{1+\cos(2\theta)}{2}$ : $= \int_0^{\pi/3} (8\frac{1+\cos(2\theta)}{2} - 2\cos\theta - 1) \, d\theta = \int_0^{\pi/3} (3 + 4\cos(2\theta) - 2\cos\theta) \, d\theta$ $= [3\theta + 2\sin(2\theta) - 2\sin\theta]_0^{\pi/3} = (3(\frac{\pi}{3}) + 2\sin(\frac{2\pi}{3}) - 2\sin(\frac{\pi}{3})) - 0 = \pi + 2(\frac{\sqrt{3}}{2}) - 2(\frac{\sqrt{3}}{2}) = \pi$
A student sets up the polar integral $\int_0^{2\pi} \int_0^{\sin\theta} r^3 \, dr \, d\theta$ but realizes the limits are incorrect because $\sin\theta$ becomes negative. What is the correct way to handle this integral?
Answer: $\int_0^{\pi} \int_0^{\sin\theta} r^3 \, dr \, d\theta$ since $\sin\theta \geq 0$ on $[0,\pi]$ When working with polar integrals, you must ensure that your radial limits make physical sense. The radius $r$ represents a distance, so it must always be non-negative. This is the key insight for handling bounds that involve trigonometric functions. The correct approach is D: $\int_0^{\pi} \int_0^{\sin\theta} r^3 \, dr \, d\theta$ . Since $\sin\theta \geq 0$ for $\theta \in [0,\pi]$ , this integral has valid bounds throughout the entire domain. The region being integrated is likely the upper half of a circle or rose petal where $r$ ranges from 0 to $\sin\theta$ . A is incorrect because doubling the integral assumes the integrand has the same sign over both halves of the region, which isn't necessarily true for $r^3$ . The symmetry argument fails here. B uses absolute value, which seems logical but actually changes the geometry of the region. When $\sin\theta < 0$ , using $|\sin\theta|$ means you're integrating over a different region than originally intended, potentially double-counting areas. C attempts to split the integral but makes an error in the second part. Using $-\sin\theta$ as an upper limit when $\sin\theta < 0$ does give a positive bound, but this approach unnecessarily complicates the problem and risks computational errors. Study tip: When you encounter negative bounds in polar coordinates, first check if restricting the angular domain to where the bound is naturally positive solves your problem. This is often the intended approach and avoids the complications of absolute values or integral splitting.
An iterated integral is given by $I = \int_{0}^{1} \int_{0}^{\sqrt{1-(x-1)^2}} \frac{1}{\sqrt{x^2+y^2}} \, dy \, dx$ . Which polar integral is equivalent to $I$ ?
Answer: $\int_{0}^{\pi/2} \int_0^{2\cos\theta} \, dr \, d\theta$ When converting double integrals from Cartesian to polar coordinates, you need to identify the region of integration and transform both the limits and the integrand appropriately. First, let's analyze the region. The limits show $0 \leq x \leq 1$ and $0 \leq y \leq \sqrt{1-(x-1)^2}$ . The upper boundary $y = \sqrt{1-(x-1)^2}$ can be rewritten as $y^2 = 1-(x-1)^2$ , or $(x-1)^2 + y^2 = 1$ . This is a circle with center at $(1,0)$ and radius 1. Since we're taking the positive square root and $x$ goes from 0 to 1, we're integrating over the upper semicircle from the y-axis to the rightmost point. In polar coordinates, this circle has equation $r = 2\cos\theta$ . To see why, substitute $x = r\cos\theta$ and $y = r\sin\theta$ into $(x-1)^2 + y^2 = 1$ : $(r\cos\theta - 1)^2 + (r\sin\theta)^2 = 1$ , which simplifies to $r = 2\cos\theta$ . The angular limits are $0 \leq \theta \leq \pi/2$ for the upper semicircle. The integrand $\frac{1}{\sqrt{x^2+y^2}} = \frac{1}{r}$ in polar coordinates, and we multiply by the Jacobian $r$ , giving us $\frac{1}{r} \cdot r = 1$ . Answer D correctly gives $\int_{0}^{\pi/2} \int_0^{2\cos\theta} \, dr \, d\theta$ . Answer A uses incorrect radial limits (should be $2\cos\theta$ , not 1). Answer B uses $2\sin\theta$ , which describes a different circle. Answer C has wrong angular limits and includes an extra $r$ factor. Study tip: Always sketch the region first, then identify key geometric features like circle centers and radii to determine the correct polar equation.
Let $R$ be the smaller region bounded by the circle $r=2$ , the line $y=x$ , and the $y$ -axis in the upper-half plane. Which of the following integrals correctly represents the area of $R$ ?
Answer: $\int_{\pi/4}^{\pi/2} \int_0^2 r \, dr \, d\theta$ The region $R$ is in the upper-half plane ( $y \ge 0$ ). It is bounded by the line $y=x$ , the $y$ -axis ( $x=0$ ), and the circle $r=2$ ( $x^2+y^2=4$ ). The $y$ -axis corresponds to $\theta = \pi/2$ . The line $y=x$ corresponds to $\theta = \pi/4$ . The region is a sector of the circle $r=2$ . The angle $\theta$ ranges from the line $y=x$ to the $y$ -axis, so $\pi/4 \le \theta \le \pi/2$ . The radius $r$ ranges from the origin to the circle, so $0 \le r \le 2$ . The area is given by $\iint_R dA = \iint_R r \, dr \, d\theta$ . Therefore, the correct setup is $\int_{\pi/4}^{\pi/2} \int_0^2 r \, dr \, d\theta$ .
What is the average value of the function $f(x,y) = y$ over the region $D$ which is the upper half of the disk $x^2+y^2 \le 4$ ?
Answer: $\frac{8}{3\pi}$ The average value of a function $f$ over a region $D$ is given by $\frac{1}{\text{Area}(D)} \iint_D f(x,y) \, dA$ . The region $D$ is a semicircle of radius 2, so its area is $\frac{1}{2}\pi(2^2) = 2\pi$ . In polar coordinates, the region is $0 \le r \le 2$ , $0 \le \theta \le \pi$ . The function is $f(x,y) = y = r\sin\theta$ . We need to compute the integral: $\iint_D y \, dA = \int_0^{\pi} \int_0^2 (r\sin\theta) \cdot r \, dr \, d\theta = \int_0^{\pi} \sin\theta \left( \int_0^2 r^2 \, dr \right) d\theta$ The inner integral is $\int_0^2 r^2 \, dr = [\frac{r^3}{3}]_0^2 = \frac{8}{3}$ . The outer integral is $\int_0^{\pi} \frac{8}{3}\sin\theta \, d\theta = \frac{8}{3}[-\cos\theta]_0^{\pi} = \frac{8}{3}(-(-1) - (-1)) = \frac{8}{3}(2) = \frac{16}{3}$ . The average value is $\frac{1}{2\pi} \cdot \frac{16}{3} = \frac{8}{3\pi}$ .
Consider the integral $I = \int_0^a \int_0^x \sqrt{x^2+y^2} \, dy \, dx$ . After converting to polar coordinates, what is the value of the inner integral with respect to $r$ ?
Answer: $\frac{a^3 \sec^3\theta}{3}$ The region of integration is a triangle in the first quadrant with vertices at $(0,0)$ , $(a,0)$ , and $(a,a)$ . The bounds are $0 \le x \le a$ and $0 \le y \le x$ . The line $y=0$ is the polar axis, $\theta=0$ . The line $y=x$ corresponds to $\theta = \pi/4$ . The vertical line $x=a$ is $r\cos\theta = a$ , or $r = a\sec\theta$ . So, the region in polar coordinates is $0 \le \theta \le \pi/4$ and $0 \le r \le a\sec\theta$ . The integrand is $\sqrt{x^2+y^2} = r$ , and $dA = r \, dr \, d\theta$ . The integral becomes: $I = \int_0^{\pi/4} \int_0^{a\sec\theta} r \cdot r \, dr \, d\theta = \int_0^{\pi/4} \left( \int_0^{a\sec\theta} r^2 \, dr \right) d\theta$ The question asks for the value of the inner integral. $\int_0^{a\sec\theta} r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^{a\sec\theta} = \frac{(a\sec\theta)^3}{3} - 0 = \frac{a^3 \sec^3\theta}{3}$
Consider the integral $\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \sqrt{x^2+y^2} \, dy \, dx$ . After converting to polar coordinates and evaluating, what is the result?
Answer: $\frac{16\pi}{3}$ When you encounter a double integral with limits that describe a circle and an integrand involving $\sqrt{x^2+y^2}$ , polar coordinates are your best tool. The region $-2 \leq x \leq 2$ and $-\sqrt{4-x^2} \leq y \leq \sqrt{4-x^2}$ describes a disk of radius 2 centered at the origin, since $x^2 + y^2 \leq 4$ . Converting to polar coordinates: $x = r\cos\theta$ , $y = r\sin\theta$ , and $dx\,dy = r\,dr\,d\theta$ . The integrand $\sqrt{x^2+y^2}$ becomes simply $r$ . The region becomes $0 \leq r \leq 2$ and $0 \leq \theta \leq 2\pi$ . The integral transforms to: $\int_0^{2\pi} \int_0^2 r \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^2 r^2 \, dr \, d\theta$ Evaluating: $\int_0^{2\pi} \left[\frac{r^3}{3}\right]_0^2 d\theta = \int_0^{2\pi} \frac{8}{3} d\theta = \frac{8}{3} \cdot 2\pi = \frac{16\pi}{3}$ Looking at the wrong answers: Choice A ( $\frac{4\pi}{3}$ ) results from forgetting the extra $r$ from the Jacobian. Choice B ( $\frac{8\pi}{3}$ ) comes from integrating over only half the circle ( $0$ to $\pi$ instead of $0$ to $2\pi$ ). Choice C ( $\frac{32\pi}{3}$ ) occurs if you incorrectly use $r^3$ as your integrand instead of $r^2$ . The answer is D. Key strategy: Always remember that converting to polar coordinates requires the Jacobian factor $r$ , making your integrand $r \cdot \sqrt{x^2+y^2} = r^2$ , not just $r$ .
The probability density function for the location of a particle on the $xy$ -plane is given by $f(x,y) = C e^{-(x^2+y^2)}$ for all $(x,y)$ . What is the probability that the particle is located within the disk $x^2+y^2 \le 4$ ?
Answer: $1 - e^{-4}$ When you encounter a probability density function over a region, you need to integrate the function over that region to find the probability. However, since this is a probability density function, you must first ensure it's properly normalized. To find the probability that the particle lies within the disk $x^2 + y^2 \leq 4$ , you need to evaluate $\int\int_{x^2+y^2 \leq 4} f(x,y) \, dA$ . Since the function and region both involve $x^2 + y^2$ , convert to polar coordinates where $x^2 + y^2 = r^2$ , $dA = r \, dr \, d\theta$ , and the disk becomes $0 \leq r \leq 2$ , $0 \leq \theta \leq 2\pi$ . The integral becomes: $\int_0^{2\pi} \int_0^2 Ce^{-r^2} \cdot r \, dr \, d\theta = 2\pi C \int_0^2 re^{-r^2} \, dr$ Using substitution $u = -r^2$ , $du = -2r \, dr$ : $2\pi C \int_0^2 re^{-r^2} \, dr = \pi C \int_0^{-4} e^u \, du = \pi C(1 - e^{-4})$ Since $f(x,y)$ is a probability density function, it must integrate to 1 over all space. By similar calculation over the entire plane (with limits $r: 0 \to \infty$ ), we get $C = \frac{1}{\pi}$ . Therefore, the probability is $\pi \cdot \frac{1}{\pi} \cdot (1 - e^{-4}) = 1 - e^{-4}$ , which is answer D. Answer A uses the wrong exponent (-2 instead of -4). Answer B gives $e^{-4}$ instead of $1 - e^{-4}$ . Answer C fails to account for the normalization constant. Key strategy: Always check if the density function is normalized, and use polar coordinates when dealing with circular regions involving $x^2 + y^2$ .
To evaluate $\iint_D xy \, dA$ where $D$ is the region in the first quadrant bounded by $x^2+y^2=1$ , $x^2+y^2=9$ , $y=x$ , and $y=x\sqrt{3}$ , a student converts to polar coordinates. What is the correct setup?
Answer: $\int_{\pi/4}^{\pi/3} \int_1^3 r^3\cos\theta\sin\theta \, dr \, d\theta$ When converting double integrals to polar coordinates, you need to carefully transform both the integrand and the region of integration. The key steps are: convert $x = r\cos\theta$ and $y = r\sin\theta$ , include the Jacobian $r$ , and determine the correct bounds. For the integrand $xy$ , substituting gives $xy = r\cos\theta \cdot r\sin\theta = r^2\cos\theta\sin\theta$ . Including the Jacobian $r$ from the coordinate transformation, the integrand becomes $r^3\cos\theta\sin\theta$ . Now for the region bounds. The circles $x^2 + y^2 = 1$ and $x^2 + y^2 = 9$ correspond to $r = 1$ and $r = 3$ . For the angular bounds, convert the lines: $y = x$ gives $\tan\theta = 1$ , so $\theta = \pi/4$ . The line $y = x\sqrt{3}$ gives $\tan\theta = \sqrt{3}$ , so $\theta = \pi/3$ . Since we're in the first quadrant moving counterclockwise from the steeper angle to the less steep angle, $\theta$ goes from $\pi/4$ to $\pi/3$ . Choice C correctly has $r^3\cos\theta\sin\theta$ with bounds $\pi/4$ to $\pi/3$ . Choice A uses $r^2$ instead of $r^3$ , missing the Jacobian factor. Choice B has the wrong angular bounds ( $\pi/6$ to $\pi/4$ ). Choice D reverses the order of the angular limits, which would give a negative result. Remember: always include the Jacobian $r$ when converting to polar coordinates, and sketch the region to verify your angle bounds match the geometry.

Start this quiz