Jacobian in Polar Coordinates - Multivariable Calculus

Let $R$ be the region in the $xy$ -plane defined by the inequality $x^2 + (y-1)^2 \le 1$ . Which of the following iterated integrals represents the volume of the solid under the surface $z = \sqrt{x^2+y^2}$ and above the region $R$ ?
Answer: $\int_{0}^{\pi} \int_{0}^{2\sin\theta} r^2 \,dr\,d\theta$ The volume is given by the double integral $\iint_R \sqrt{x^2+y^2} \,dA$ . To evaluate this, we convert to polar coordinates. The region $R$ is a circle of radius 1 centered at $(0,1)$ . The inequality $x^2 + y^2 - 2y + 1 \le 1$ becomes $x^2 + y^2 \le 2y$ . In polar coordinates, this is $r^2 \le 2r\sin\theta$ , which simplifies to $r \le 2\sin\theta$ . For the circle to be traced completely, $\theta$ must range from $0$ to $\pi$ . The integrand $z = \sqrt{x^2+y^2}$ becomes $z=r$ . The area element is $dA = r\,dr\,d\theta$ . Therefore, the integral is $\int_{0}^{\pi} \int_{0}^{2\sin\theta} r \cdot r \,dr\,d\theta = \int_{0}^{\pi} \int_{0}^{2\sin\theta} r^2 \,dr\,d\theta$ .
Which of the following polar integrals is equivalent to the Cartesian integral $\int_{0}^{2} \int_{x}^{\sqrt{8-x^2}} \frac{1}{5+x^2+y^2} \,dy\,dx$ ?
Answer: $\int_{\pi/4}^{\pi/2} \int_{0}^{\sqrt{8}} \frac{r}{5+r^2} \,dr\,d\theta$ When converting double integrals from Cartesian to polar coordinates, you need to carefully analyze the region of integration and transform both the integrand and the area element. First, let's identify the region. The bounds are $0 \leq x \leq 2$ and $x \leq y \leq \sqrt{8-x^2}$ . The upper boundary $y = \sqrt{8-x^2}$ represents the upper semicircle $x^2 + y^2 = 8$ , while the lower boundary $y = x$ is the line making a 45° angle with the x-axis. This creates a region between the line $y = x$ (which corresponds to $\theta = \pi/4$ ) and the positive y-axis ( $\theta = \pi/2$ ), bounded by the circle of radius $\sqrt{8}$ . In polar coordinates, $x^2 + y^2 = r^2$ , so the integrand $\frac{1}{5+x^2+y^2}$ becomes $\frac{1}{5+r^2}$ . The area element $dy\,dx$ transforms to $r\,dr\,d\theta$ , giving us $\frac{r}{5+r^2}$ in the integrand. The correct bounds are $\pi/4 \leq \theta \leq \pi/2$ and $0 \leq r \leq \sqrt{8}$ , making answer C correct. Answer A has the wrong $\theta$ bounds ( $0$ to $\pi/4$ instead of $\pi/4$ to $\pi/2$ ). Answer B is missing the $r$ factor in the integrand that comes from the Jacobian transformation. Answer D uses $2\sec\theta$ as the upper $r$ bound, which would represent the line $x = 2$ rather than the circular boundary. Remember: always include the Jacobian factor $r$ when converting to polar coordinates, and sketch the region to determine the correct angular bounds.
For the integral $\int_0^{2\pi} \int_0^a g(r) \, r \, dr \, d\theta = \pi a^2$ , what can you deduce about the function $g(r)$ and the role of the Jacobian in this result?
Answer: $g(r) = 1$ and the Jacobian $r$ gives the result $\pi a^2$ , which equals the area of a disk of radius $a$ When you encounter a double integral in polar coordinates, always remember that the Jacobian factor $r$ appears naturally from the coordinate transformation, and the limits tell you about the geometric region being integrated. Let's work through this systematically. The integral $\int_0^{2\pi} \int_0^a g(r) \cdot r \, dr \, d\theta = \pi a^2$ describes integration over a disk of radius $a$ . If $g(r) = 1$ , then we have: $\int_0^{2\pi} \int_0^a 1 \cdot r \, dr \, d\theta = \int_0^{2\pi} \left[\frac{r^2}{2}\right]_0^a d\theta = \int_0^{2\pi} \frac{a^2}{2} d\theta = \frac{a^2}{2} \cdot 2\pi = \pi a^2$ This matches our given result perfectly, confirming that $g(r) = 1$ and the Jacobian $r$ produces the familiar disk area formula. Answer A incorrectly suggests $g(r) = \frac{1}{r}$ , but this would create $\int r \cdot \frac{1}{r} dr = \int 1 dr = a$ , not $\frac{a^2}{2}$ . Answer B proposes $g(r) = r$ , which would give $\int r^2 dr = \frac{a^3}{3}$ , leading to $\frac{2\pi a^3}{3}$ , not $\pi a^2$ . Answer D suggests $g(r) = \frac{1}{2\pi}$ , but this constant would yield $\frac{a^2}{2}$ , missing the $\pi$ factor entirely. Study tip: In polar coordinate problems, when the result matches a familiar geometric formula (like disk area), check if $g(r) = 1$ first—the Jacobian often provides exactly the weighting needed for standard geometric results.
Consider the double integral $\iint_D \frac{y}{x^2 + y^2} \, dA$ where $D$ is the sector $\{(r,\theta): 1 \leq r \leq 3, 0 \leq \theta \leq \pi/3\}$ in polar coordinates. After applying the transformation and Jacobian, what is the resulting integrand?
Answer: $\frac{r\sin\theta}{r^2} \cdot r = \sin\theta$ When converting double integrals from rectangular to polar coordinates, you need to transform both the integrand and the differential area element. The key steps are: substitute $x = r\cos\theta$ and $y = r\sin\theta$ , then multiply by the Jacobian $r$ . Let's work through this systematically. Starting with $\frac{y}{x^2 + y^2}$ , we substitute the polar coordinate relationships. The numerator becomes $y = r\sin\theta$ . For the denominator, we have $x^2 + y^2 = (r\cos\theta)^2 + (r\sin\theta)^2 = r^2(\cos^2\theta + \sin^2\theta) = r^2$ . So our integrand transforms to $\frac{r\sin\theta}{r^2} = \frac{\sin\theta}{r}$ . Finally, we multiply by the Jacobian $r$ to get $\frac{\sin\theta}{r} \cdot r = \sin\theta$ . This confirms answer D is correct. Looking at the wrong answers: Choice A incorrectly shows the denominator as just $r$ instead of $r^2$ , missing that $x^2 + y^2 = r^2$ . Choice B makes the same denominator error as A, then mysteriously multiplies by $r^2$ instead of the correct Jacobian $r$ . Choice C correctly transforms the integrand to $\frac{\sin\theta}{r}$ but fails to apply the Jacobian, stopping short of the complete conversion. Remember: when converting to polar coordinates, always check that you've applied both the coordinate substitution AND the Jacobian. The Jacobian $r$ often simplifies expressions beautifully, as it does here by eliminating the $\frac{1}{r}$ factor.
When evaluating $\iint_D \frac{1}{\sqrt{x^2 + y^2}} \, dA$ where $D$ is the annulus $1 \leq x^2 + y^2 \leq 9$ , a student transforms to polar coordinates and writes $\int_0^{2\pi} \int_1^3 \frac{1}{r} \cdot r \, dr \, d\theta$ . What is the most likely reasoning behind their approach?
Answer: They recognized that $\sqrt{x^2 + y^2} = r$ and correctly applied the Jacobian $r$ to get $\frac{r}{r} = 1$ The student correctly identified that $\sqrt{x^2 + y^2} = r$ in polar coordinates, so $\frac{1}{\sqrt{x^2 + y^2}} = \frac{1}{r}$ . They then correctly applied the Jacobian $r$ , giving $\frac{1}{r} \cdot r = 1$ . The bounds are also correct since $1 \leq x^2 + y^2 \leq 9$ becomes $1 \leq r^2 \leq 9$ or $1 \leq r \leq 3$ . Choices B, C, and D suggest errors where there are none.
Using the information in the passage, what is the value of $I = \int_{-\infty}^{\infty} e^{-x^2} \,dx$ ?
Answer: $\sqrt{\pi}$ We compute $I^2$ by converting the double integral over the entire plane $\mathbb{R}^2$ to polar coordinates. The region is $0 \le r < \infty$ and $0 \le \theta \le 2\pi$ . The integrand $e^{-(x^2+y^2)}$ becomes $e^{-r^2}$ . The area element is $dA = r\,dr\,d\theta$ . So, $I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r \,dr\,d\theta$ . We evaluate the inner improper integral using a substitution $u = -r^2$ , so $du = -2r\,dr$ . $\int_{0}^{\infty} e^{-r^2} r \,dr = \lim_{b\to\infty} \int_{0}^{b} e^{-r^2} r \,dr = \lim_{b\to\infty} \left[-\frac{1}{2}e^{-r^2}\right]_0^b = \lim_{b\to\infty} \left(-\frac{1}{2}e^{-b^2} + \frac{1}{2}e^0\right) = 0 + \frac{1}{2} = \frac{1}{2}$ . Now, we evaluate the outer integral: $I^2 = \int_{0}^{2\pi} \frac{1}{2} \,d\theta = \frac{1}{2} [\theta]_0^{2\pi} = \frac{1}{2}(2\pi) = \pi$ . Since $I^2 = \pi$ and the integrand $e^{-x^2}$ is always positive, $I$ must be positive. Therefore, $I = \sqrt{\pi}$ .
When converting a double integral from Cartesian coordinates $(x,y)$ to polar coordinates $(r, \theta)$ , the differential area element $dA = dx\,dy$ is replaced by $r\,dr\,d\theta$ . What is the primary geometric reason for the inclusion of the factor $r$ ?
Answer: The factor $r$ reflects that the area of a polar grid element increases as its distance from the origin increases. A small region in polar coordinates, a 'polar rectangle', is defined by $r \le r^* \le r+dr$ and $\theta \le \theta^* \le \theta+d\theta$ . This region is approximately a trapezoid or a rectangle. Its sides are of length $dr$ (in the radial direction) and $r\,d\theta$ (the arc length). The area of this element is approximately $(dr)(r\,d\theta) = r\,dr\,d\theta$ . The factor $r$ is crucial because the arc length corresponding to a given $d\theta$ is proportional to the radius $r$ . Thus, polar grid 'rectangles' farther from the origin have a larger area than those closer to the origin, for the same $dr$ and $d\theta$ .
A double integral is set up in polar coordinates to find the area of the region bounded by one loop of the rose curve $r = 3\cos(2\theta)$ . The calculation begins with $\int_{-\pi/4}^{\pi/4} \left( \int_{0}^{3\cos(2\theta)} r \,dr \right) \,d\theta$ . What is the value of this area?
Answer: $\frac{9\pi}{8}$ The area is given by the integral $A = \iint_R dA = \int_{-\pi/4}^{\pi/4} \int_{0}^{3\cos(2\theta)} r \,dr\,d\theta$ . First, evaluate the inner integral: $\int_{0}^{3\cos(2\theta)} r \,dr = \left[\frac{r^2}{2}\right]_0^{3\cos(2\theta)} = \frac{9\cos^2(2\theta)}{2}$ . Now, evaluate the outer integral: $A = \int_{-\pi/4}^{\pi/4} \frac{9\cos^2(2\theta)}{2} \,d\theta$ . Using the identity $\cos^2(x) = \frac{1+\cos(2x)}{2}$ , we get $A = \frac{9}{2} \int_{-\pi/4}^{\pi/4} \frac{1+\cos(4\theta)}{2} \,d\theta = \frac{9}{4} \left[\theta + \frac{1}{4}\sin(4\theta)\right]_{-\pi/4}^{\pi/4}$ . Evaluating at the limits: $\frac{9}{4} \left[ (\frac{\pi}{4} + \frac{1}{4}\sin(\pi)) - (-\frac{\pi}{4} + \frac{1}{4}\sin(-\pi)) \right] = \frac{9}{4} (\frac{\pi}{4} + \frac{\pi}{4}) = \frac{9}{4} (\frac{\pi}{2}) = \frac{9\pi}{8}$ .
The area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ can be found using the change of variables $x=ar\cos\theta$ and $y=br\sin\theta$ . What is the area of the ellipse $\frac{x^2}{16} + \frac{y^2}{25} = 1$ , calculated by evaluating $\iint_R dA$ over the elliptical region $R$ with this transformation?
Answer: $20\pi$ When you encounter an ellipse area problem using change of variables, you're dealing with a double integral transformation where the Jacobian determines how area elements transform between coordinate systems. For the transformation $x = ar\cos\theta$ and $y = br\sin\theta$ , you need to find the Jacobian determinant. Computing the partial derivatives: $\frac{\partial x}{\partial r} = a\cos\theta$ , $\frac{\partial x}{\partial \theta} = -ar\sin\theta$ , $\frac{\partial y}{\partial r} = b\sin\theta$ , and $\frac{\partial y}{\partial \theta} = br\cos\theta$ . The Jacobian is: $J = \begin{vmatrix} a\cos\theta & -ar\sin\theta \\ b\sin\theta & br\cos\theta \end{vmatrix} = abr\cos^2\theta + abr\sin^2\theta = abr$ For your ellipse $\frac{x^2}{16} + \frac{y^2}{25} = 1$ , we have $a^2 = 16$ so $a = 4$ , and $b^2 = 25$ so $b = 5$ . The ellipse transforms to the unit disk where $0 \leq r \leq 1$ and $0 \leq \theta \leq 2\pi$ . The area integral becomes: $\iint_R dA = \int_0^{2\pi}\int_0^1 |J| \, dr \, d\theta = \int_0^{2\pi}\int_0^1 abr \, dr \, d\theta = ab\int_0^{2\pi}\int_0^1 r \, dr \, d\theta = ab \cdot 2\pi \cdot \frac{1}{2} = \pi ab$ Therefore: $\pi ab = \pi(4)(5) = 20\pi$ . Answer A ( $9\pi$ ) likely comes from using $(a-b)\pi = \pi$ . Answer B ( $\pi$ ) ignores the scaling factors entirely. Answer D ( $40\pi$ ) incorrectly uses $2\pi ab$ instead of $\pi ab$ . Remember: when transforming areas, always include the absolute value of the Jacobian determinant, and for ellipses, the general area formula is $\pi ab$ where $a$ and $b$ are the semi-axes lengths.
Let $R$ be the region in the right half-plane ( $x \ge 0$ ) lying between the cardioid $r = 1 + \cos\theta$ and the circle $r=1$ . Which integral represents the area of $R$ ?
Answer: $\int_{-\pi/2}^{\pi/2} \int_{1}^{1+\cos\theta} r \,dr\,d\theta$ When setting up double integrals in polar coordinates to find area, you need to remember two key components: the correct bounds of integration and the proper area element, which is $r \, dr \, d\theta$ (not just $dr \, d\theta$ ). For this problem, you're finding the area between two curves in the right half-plane. The region $R$ is bounded by the inner circle $r = 1$ and outer cardioid $r = 1 + \cos\theta$ . Since you want only the right half-plane ( $x \geq 0$ ), you need $\theta$ to range from $-\pi/2$ to $\pi/2$ . For each fixed angle $\theta$ in this range, $r$ varies from the inner boundary $r = 1$ to the outer boundary $r = 1 + \cos\theta$ . Answer B correctly captures both requirements: $\int_{-\pi/2}^{\pi/2} \int_{1}^{1+\cos\theta} r \,dr\,d\theta$ Answer A uses the wrong area element ( $1$ instead of $r$ ) – this would give you area in rectangular coordinates, not polar. Answer C has the correct area element but wrong $\theta$ bounds ( $0$ to $2\pi$ instead of $-\pi/2$ to $\pi/2$ ), which would give you the entire region around both curves, not just the right half-plane. Answer D has the wrong inner $r$ -bound (starting from $0$ instead of $1$ ), which would include the area inside the circle $r = 1$ rather than just the region between the two curves. Remember: in polar coordinates, area integrals always need the factor $r$ in the integrand, and carefully determine your bounds by visualizing the region you want.
A student evaluating $\iint_R xy \, dA$ over the region $R = \{(x,y): x^2 + y^2 \leq 4, x \geq 0, y \geq 0\}$ transforms to polar coordinates and obtains $\int_0^{\pi/2} \int_0^2 r^2 \cos\theta \sin\theta \cdot r \, dr \, d\theta$ . Which aspect of their work demonstrates correct understanding of the Jacobian?
Answer: They correctly identified that $xy = r^2\cos\theta\sin\theta$ and included the necessary factor $r$ from the Jacobian When transforming double integrals from Cartesian to polar coordinates, you must account for both the coordinate transformation and the Jacobian factor that adjusts for how area elements change between coordinate systems. The student correctly identified that $xy = r\cos\theta \cdot r\sin\theta = r^2\cos\theta\sin\theta$ when converting the integrand to polar coordinates. Crucially, they also included the additional factor $r$ from the Jacobian of the polar transformation. The Jacobian for polar coordinates is $r$ , which means $dA = dx\,dy$ becomes $r\,dr\,d\theta$ . This explains why their final integrand is $r^2\cos\theta\sin\theta \cdot r = r^3\cos\theta\sin\theta$ , showing they understand both parts of the transformation. Looking at the wrong answers: (A) confuses changing limits of integration with applying the Jacobian—these are separate steps in coordinate transformation. (B) incorrectly states the Jacobian as $r^2$ when it's actually $r$ , and the $r^2$ term comes from the $xy$ transformation. (C) suggests the Jacobian changes the integrand $xy$ itself, but the Jacobian only provides an additional multiplicative factor—the integrand transforms according to the coordinate relationships. Study tip: Remember that coordinate transformations involve two distinct steps: (1) convert the integrand using coordinate relationships ( $x = r\cos\theta$ , $y = r\sin\theta$ ), and (2) multiply by the Jacobian factor ( $r$ for polar coordinates). The Jacobian doesn't change what the original function becomes—it's an additional factor that accounts for how area elements stretch or compress.
A student sets up the integral $\int_0^{2\pi} \int_0^3 f(r,\theta) \cdot r^2 \, dr \, d\theta$ claiming they used the correct Jacobian for polar coordinates. What can you conclude about their work?
Answer: They made an error because the Jacobian for polar coordinates is $r$ , not $r^2$ , regardless of the integrand The Jacobian for the transformation from Cartesian to polar coordinates is always $r$ , so $dA = r \, dr \, d\theta$ . Using $r^2$ as a factor suggests the student confused the Jacobian with something else, possibly the integrand. Choice A incorrectly suggests the Jacobian can change based on the integrand. Choice C introduces cylindrical coordinates, which is not relevant here. Choice D focuses on limits rather than the Jacobian error.
A student claims that for the transformation $x = r\cos\theta, y = r\sin\theta$ , the Jacobian can be computed as $J = \frac{\partial x}{\partial r} \cdot \frac{\partial y}{\partial \theta} = \cos\theta \cdot r\cos\theta = r\cos^2\theta$ . What is wrong with this reasoning?
Answer: The student computed the Jacobian determinant incorrectly by multiplying diagonal elements instead of using the proper formula The Jacobian determinant is $\begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r(\cos^2\theta + \sin^2\theta) = r$ . The student only multiplied two elements instead of computing the full determinant. Choice A is incorrect because they did use partial derivatives. Choice B is wrong about the derivative of $y$ . Choice D suggests computing the wrong transformation.
When transforming the double integral $\iint_R (x^2 + y^2) \, dA$ over the region $R = \{(x,y): x^2 + y^2 \leq 4, x \geq 0, y \geq 0\}$ to polar coordinates, which expression correctly includes the Jacobian?
Answer: $\int_0^{\pi/2} \int_0^2 r^2 \cdot r \, dr \, d\theta$ The Jacobian for polar coordinates is $r$ , so $dA = r \, dr \, d\theta$ . Since $x^2 + y^2 = r^2$ in polar coordinates, the integral becomes $\int_0^{\pi/2} \int_0^2 r^2 \cdot r \, dr \, d\theta$ . Choice B uses the incorrect Jacobian $1/r$ . Choice C omits the Jacobian entirely. Choice D incorrectly uses $r^2$ as the Jacobian.
Consider the transformation from Cartesian to polar coordinates: $x = r\cos\theta, y = r\sin\theta$ . If a student incorrectly computes the Jacobian determinant as $\frac{\partial(x,y)}{\partial(r,\theta)} = r\cos\theta + r\sin\theta$ , what fundamental error did they make?
Answer: They computed the trace of the Jacobian matrix instead of computing the determinant properly The Jacobian determinant should be $\begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r$ . The student computed $r\cos\theta + r\sin\theta$ , which is the trace (sum of diagonal elements) rather than the determinant. Choice B is incorrect because the signs are handled properly in the determinant calculation. Choice C describes a different error. Choice D is too vague.
When setting up $\iint_D f(x,y) \, dA$ in polar coordinates where $D$ is defined by $1 \leq r \leq 2$ and $0 \leq \theta \leq \pi/4$ , a student writes $\int_0^{\pi/4} \int_1^2 f(r\cos\theta, r\sin\theta) \, dr \, d\theta$ . What critical component is missing?
Answer: The Jacobian $r$ must be included as a factor in the integrand for the area element When transforming to polar coordinates, the area element $dA = dx \, dy$ becomes $dA = r \, dr \, d\theta$ due to the Jacobian. The student forgot to include the factor $r$ . The correct integral should be $\int_0^{\pi/4} \int_1^2 f(r\cos\theta, r\sin\theta) \cdot r \, dr \, d\theta$ . Choice A is incorrect about order of integration. Choice C misunderstands function notation. Choice D is wrong about the bounds.
Evaluate the integral $\int_{-3}^{3} \int_{0}^{\sqrt{9-x^2}} (x^2+y^2)^{3/2} \,dy\,dx$ .
Answer: $\frac{243\pi}{5}$ The region of integration is the upper semi-disk of radius 3 centered at the origin, described by $-3 \le x \le 3$ and $0 \le y \le \sqrt{9-x^2}$ . In polar coordinates, this region is $0 \le r \le 3$ and $0 \le \theta \le \pi$ . The integrand $(x^2+y^2)^{3/2}$ becomes $(r^2)^{3/2} = r^3$ . The area element is $dA = r\,dr\,d\theta$ . The integral becomes $\int_{0}^{\pi} \int_{0}^{3} (r^3) \cdot r \,dr\,d\theta = \int_{0}^{\pi} \int_{0}^{3} r^4 \,dr\,d\theta$ . The inner integral is $\int_{0}^{3} r^4 \,dr = \left[\frac{r^5}{5}\right]_0^3 = \frac{243}{5}$ . The outer integral is $\int_{0}^{\pi} \frac{243}{5} \,d\theta = \frac{243}{5}[\theta]_0^\pi = \frac{243\pi}{5}$ .
A thin plate is in the shape of the region in the first quadrant bounded by the circle $x^2+y^2=4$ and the line $y=x$ . The density of the plate at any point $(x,y)$ is given by $\delta(x,y) = x$ . Which of the following integrals represents the mass of the plate?
Answer: $\int_{0}^{\pi/4} \int_{0}^{2} r^2 \cos\theta \,dr\,d\theta$ When finding the mass of a plate with variable density, you need to set up a double integral of the density function over the region. Since this problem involves a circular boundary, polar coordinates will simplify the integration. First, identify the region: it's bounded by the circle $x^2 + y^2 = 4$ (radius 2) and the line $y = x$ in the first quadrant. In polar coordinates, the circle becomes $r = 2$ and the line $y = x$ corresponds to $\theta = \pi/4$ (since $\tan\theta = y/x = 1$ ). So the region spans from $\theta = 0$ to $\theta = \pi/4$ and $r = 0$ to $r = 2$ . The density function is $\delta(x,y) = x$ . Converting to polar coordinates: $x = r\cos\theta$ , so $\delta = r\cos\theta$ . The mass integral becomes $\iint_R \delta(x,y) \, dA = \int_0^{\pi/4} \int_0^2 r\cos\theta \cdot r \, dr \, d\theta = \int_0^{\pi/4} \int_0^2 r^2\cos\theta \, dr \, d\theta$ . This matches choice D. Choice A uses the wrong radius limit (4 instead of 2) — remember that $x^2 + y^2 = 4$ gives $r = 2$ , not $r = 4$ . Choice B has the correct limits but is missing the extra factor of $r$ from the Jacobian of the polar coordinate transformation. Choice C uses the wrong angular limit ( $\pi/2$ instead of $\pi/4$ ) — this would include the entire first quadrant rather than just the region below the line $y = x$ . Remember: in polar coordinates, always include the Jacobian factor $r$ in your integrand, and carefully convert both the region boundaries and the function to polar form.
Consider the integral $\iint_R e^{x^2 + y^2} \, dA$ over the region $R = \{(x,y): x^2 + y^2 \leq 1\}$ . When transformed to polar coordinates, what role does the Jacobian play in the resulting integrand?
Answer: The Jacobian $r$ multiplies $e^{r^2}$ to give $re^{r^2}$ , which has a convenient antiderivative In polar coordinates, $e^{x^2 + y^2} = e^{r^2}$ and the Jacobian is $r$ , so the integrand becomes $re^{r^2}$ . This is significant because $\int re^{r^2} \, dr = \frac{1}{2}e^{r^2} + C$ using the substitution $u = r^2$ . Choice B incorrectly suggests cancellation. Choice C wrongly claims the exponent becomes $r^3$ . Choice D incorrectly places $r$ in the denominator.

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