Double Integrals in Polar - Calculus 3

1. A double integral in Cartesian coordinates has the form $\int_0^a \int_0^{\sqrt{a^2-x^2}} g(x^2+y^2) \, dy \, dx$ where $a > 0$. After converting to polar coordinates and evaluating, the result is $\frac{\pi}{2}\int_0^a r \cdot g(r^2) \, dr$. What relationship must hold between the original integral and the converted form?
   Answer: The conversion is correct as stated, with no additional conditions needed When you encounter double integrals with limits involving expressions like $\sqrt{a^2-x^2}$, you're looking at a region that's naturally suited for polar coordinates. The key insight is understanding what happens during the coordinate transformation. Let's trace through this conversion step by step. The original integral $\int_0^a \int_0^{\sqrt{a^2-x^2}} g(x^2+y^2) \, dy \, dx$ describes integration over a quarter-circle of radius $a$ in the first quadrant (since $y = \sqrt{a^2-x^2}$ is the upper half of $x^2 + y^2 = a^2$). In polar coordinates, this region becomes $0 \leq r \leq a$ and $0 \leq \theta \leq \frac{\pi}{2}$. The function $g(x^2+y^2)$ naturally becomes $g(r^2)$, and we include the Jacobian $r$. The integral becomes $\int_0^{\pi/2} \int_0^a g(r^2) \cdot r \, dr \, d\theta$. Since $g(r^2)$ doesn't depend on $\theta$, we can factor out the $\theta$ integration: $\int_0^{\pi/2} d\theta \int_0^a r \cdot g(r^2) \, dr = \frac{\pi}{2}\int_0^a r \cdot g(r^2) \, dr$. Option A is wrong because $g(x^2+y^2)$ being independent of $\theta$ is automatic—it's a consequence of the function's form, not an additional requirement. Option B is incorrect because we need no special properties of $g(u)$ beyond what's given. Option D is wrong because the factor $\frac{\pi}{2}$ correctly accounts for the $\theta$ integration. Study tip: When converting to polar coordinates, always check that your new limits match the original region's geometry, and remember that radially symmetric integrands automatically satisfy any $\theta$-independence requirements.
2. A region $R$ is described in polar coordinates as $\{(r,\theta) : 0 \leq \theta \leq \pi/2, \, \theta \leq r \leq \pi - \theta\}$. When computing $\iint_R r \, dA$, what is the correct integral setup and what geometric shape does this region represent?
   Answer: $\int_0^{\pi/2} \int_{\theta}^{\pi - \theta} r^2 \, dr \, d\theta$; the region is bounded by two spirals When working with double integrals in polar coordinates, you need to carefully set up both the integral bounds and the differential area element. The key is remembering that $dA = r \, dr \, d\theta$ in polar coordinates, so when integrating a function $f(r,\theta)$, you compute $\iint_R f(r,\theta) \cdot r \, dr \, d\theta$. For the integral $\iint_R r \, dA$, you're integrating the function $f(r,\theta) = r$. This becomes $\iint_R r \cdot r \, dr \, d\theta = \iint_R r^2 \, dr \, d\theta$. The bounds come directly from the region description: $\theta$ ranges from $0$ to $\pi/2$, and for each $\theta$, $r$ ranges from $\theta$ to $\pi - \theta$. This gives us $\int_0^{\pi/2} \int_{\theta}^{\pi - \theta} r^2 \, dr \, d\theta$. To understand the geometry, note that the boundaries are $r = \theta$ and $r = \pi - \theta$. These aren't straight lines through the origin (which would have the form $\theta = \text{constant}$), but rather curves where the radial distance depends on the angle. These are spiral-like curves that create a region between them. Answer choice A has the correct integral setup but misidentifies the geometric shape. Choice C correctly identifies the spiral boundaries but incorrectly uses $r$ instead of $r^2$ in the integrand, forgetting the $r$ factor from $dA$. Choice D has wrong $\theta$ bounds and misidentifies the geometry. Study tip: Always remember that $dA = r \, dr \, d\theta$ in polar coordinates, and distinguish between radial curves ($r = g(\theta)$) and angular lines ($\theta = \text{constant}$) when visualizing regions.
3. The integral $\int_0^{2\pi} \int_1^{3} r^2 \cos(r) \, dr \, d\theta$ can be evaluated by recognizing that one part factors out. However, if the region of integration were changed to $\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}$ and $\sec\theta \leq r \leq 3$, what would be the new integral setup?
   Answer: $\int_{\pi/6}^{\pi/3} \int_{\sec\theta}^3 r^2 \cos(r) \, dr \, d\theta$ When you encounter double integrals in polar coordinates, pay careful attention to how the region of integration is described and how it translates into integral bounds. The original integral uses a simple rectangular region in polar coordinates: $r$ goes from 1 to 3, and $\theta$ goes from 0 to $2\pi$. The new region has $\theta$ bounded by $\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}$ and $r$ bounded by $\sec\theta \leq r \leq 3$. The key insight is recognizing that $\sec\theta = \frac{1}{\cos\theta}$, which represents the distance from the origin to a vertical line in Cartesian coordinates. When setting up the integral, you simply use the bounds as given: the outer integral has limits $\frac{\pi}{6}$ to $\frac{\pi}{3}$ for $\theta$, and the inner integral has limits $\sec\theta$ to $3$ for $r$. The integrand $r^2\cos(r)$ remains unchanged because we're integrating over the same type of region, just with different boundaries. Looking at the wrong answers: (A) incorrectly uses $\cos\theta$ instead of $\sec\theta = \frac{1}{\cos\theta}$. (B) writes out $\frac{1}{\cos\theta}$ explicitly, which is mathematically equivalent to $\sec\theta$ but unnecessarily complicated. (D) incorrectly includes an extra $\cos\theta$ factor in the integrand, which would change the function being integrated entirely. Study tip: When converting between polar coordinate descriptions, remember that $\sec\theta = \frac{1}{\cos\theta}$ and always preserve the original integrand unless you're changing coordinate systems entirely.
4. A region $R$ in the first quadrant is bounded by $y = x$, $y = 0$, and $x^2 + y^2 = 4$. When setting up $\iint_R f(x,y) \, dA$ in polar coordinates, which description correctly identifies the region's boundaries?
   Answer: $0 \leq \theta \leq \pi/4$ and $0 \leq r \leq 2$ When converting double integrals to polar coordinates, you need to carefully analyze how each boundary curve translates from Cartesian to polar form, then determine the proper order of integration. Let's examine each boundary of region $R$. The curve $y = 0$ corresponds to the positive $x$-axis, giving $\theta = 0$. The line $y = x$ has slope 1, so $\tan\theta = 1$, which means $\theta = \pi/4$. The circle $x^2 + y^2 = 4$ becomes $r^2 = 4$, or $r = 2$. Since we're in the first quadrant bounded by these three curves, $\theta$ ranges from $0$ to $\pi/4$, and $r$ extends from the origin ($r = 0$) to the circle ($r = 2$). Answer D correctly captures this: $0 \leq \theta \leq \pi/4$ and $0 \leq r \leq 2$. Answer A incorrectly suggests $r$ starts at $\frac{2}{\cos\theta + \sin\theta}$ rather than 0. This expression would represent some inner boundary that doesn't exist in our region. Answer B limits $r$ to $2/\cos\theta$, which represents the vertical line $x = 2$, not our actual boundary. Answer C extends $\theta$ to $\pi/2$, which would include the entire first quadrant rather than just the triangular sector below $y = x$. Study tip: Always sketch the region first, then systematically convert each boundary curve to polar coordinates. Pay special attention to whether $r$ starts from the origin or from an inner boundary—this is a common source of errors in polar integration setup.
5. The double integral $\int_0^2 \int_0^{\sqrt{4-x^2}} xy \, dy \, dx$ is converted to polar coordinates. Which of the following represents the equivalent integral?
   Answer: $\int_0^{\pi/2} \int_0^2 r^3 \cos\theta \sin\theta \, dr \, d\theta$ The region of integration is the first quadrant portion of the disk $x^2 + y^2 \leq 4$, so $0 \leq \theta \leq \pi/2$ and $0 \leq r \leq 2$. Converting: $x = r\cos\theta$, $y = r\sin\theta$, $dx\,dy = r\,dr\,d\theta$. The integrand becomes $xy \cdot r = (r\cos\theta)(r\sin\theta) \cdot r = r^3\cos\theta\sin\theta$. Choice B forgets the Jacobian factor $r$ in the integrand. Choice C uses incorrect $\theta$-limits (should be $\pi/2$, not $\pi$). Choice D uses incorrect $r$-limits (radius is 2, not 4).
6. Consider the integral $\iint_R e^{x^2+y^2} \, dA$ where $R$ is the annular region $1 \leq x^2 + y^2 \leq 9$. After converting to polar coordinates, the integral becomes $\int_0^{2\pi} \int_1^3 re^{r^2} \, dr \, d\theta$. What is the value of this integral?
   Answer: $\pi(e^9 - e)$ The integral separates as $\int_0^{2\pi} d\theta \int_1^3 re^{r^2} \, dr = 2\pi \int_1^3 re^{r^2} \, dr$. For the $r$-integral, use substitution $u = r^2$, so $du = 2r\,dr$ and $r\,dr = \frac{1}{2}du$. When $r = 1$, $u = 1$; when $r = 3$, $u = 9$. Thus $\int_1^3 re^{r^2} \, dr = \frac{1}{2}\int_1^9 e^u \, du = \frac{1}{2}(e^9 - e^1)$. The full integral is $2\pi \cdot \frac{1}{2}(e^9 - e) = \pi(e^9 - e)$. Choice B misses the $\frac{1}{2}$ from substitution. Choice C incorrectly divides by 2. Choice D forgets the $2\pi$ factor.
7. The integral $\int_0^{\pi/3} \int_0^{4\cos\theta} r^3 \sin\theta \, dr \, d\theta$ represents the evaluation of $\iint_R y \cdot f(x,y) \, dA$ in polar coordinates for some function $f$. What is $f(x,y)$?
   Answer: $f(x,y) = x^2 + y^2$ When converting double integrals from rectangular to polar coordinates, you need to recognize how the integrand transforms. The key is understanding that $dA = dx \, dy$ becomes $r \, dr \, d\theta$, and rectangular coordinates become $x = r\cos\theta$ and $y = r\sin\theta$. Looking at the given integral $\int_0^{\pi/3} \int_0^{4\cos\theta} r^3 \sin\theta \, dr \, d\theta$, you can identify the components. The integrand is $r^3 \sin\theta$, which includes the required $r$ from the area element $r \, dr \, d\theta$. This means the actual function being integrated is $r^2 \sin\theta \cdot y$, where the $y$ factor comes from the problem statement saying we're evaluating $\iint_R y \cdot f(x,y) \, dA$. Since $y = r\sin\theta$ and we have $r^2 \sin\theta$ remaining after accounting for the $y$ factor and area element, we get $r^2 = x^2 + y^2$. Therefore, $f(x,y) = x^2 + y^2$. Answer choice A, $f(x,y) = (x^2 + y^2)^{3/2}/y$, would give $r^3/\sin\theta$ in polar form, which doesn't match. Answer B, $f(x,y) = \sqrt{x^2 + y^2}$, converts to just $r$, missing the $r^2$ we need. Answer C, $f(x,y) = x^2 + y^2 - y$, would produce $r^2 - r\sin\theta$, which is more complex than our integrand. Study tip: When working backwards from polar integrals, always separate out the $r$ from $dA$ first, then convert the remaining expression using $r^2 = x^2 + y^2$ and the basic polar relationships.
8. Consider evaluating $\iint_R (x^2 - y^2) \, dA$ over the region $R$ bounded by $r = 2 + \cos\theta$. In polar coordinates, the integrand $x^2 - y^2$ becomes $r^2\cos(2\theta)$. What is the complete setup for this integral?
   Answer: $\int_0^{2\pi} \int_0^{2+\cos\theta} r^3 \cos(2\theta) \, dr \, d\theta$ When converting double integrals from Cartesian to polar coordinates, you need to handle three key components: the integrand transformation, the region bounds, and the crucial Jacobian factor. The integrand $x^2 - y^2$ correctly transforms to $r^2\cos(2\theta)$ using the identity $x^2 - y^2 = r^2(\cos^2\theta - \sin^2\theta) = r^2\cos(2\theta)$. The region $r = 2 + \cos\theta$ is a limaçon that completes one full loop as $\theta$ goes from $0$ to $2\pi$, with $r$ ranging from $0$ to $2 + \cos\theta$. The critical insight is the Jacobian factor. When converting $dA = dx \, dy$ to polar coordinates, you must include $r$ in the area element: $dA = r \, dr \, d\theta$. This transforms our integrand from $r^2\cos(2\theta)$ to $r^2\cos(2\theta) \cdot r = r^3\cos(2\theta)$, making the complete integral $\int_0^{2\pi} \int_0^{2+\cos\theta} r^3 \cos(2\theta) \, dr \, d\theta$. Choice A omits the Jacobian factor, keeping only $r^2\cos(2\theta)$ instead of $r^3\cos(2\theta)$. Choice C uses the correct integrand but wrong $\theta$ bounds—$0$ to $\pi$ would give only half the limaçon. Choice D has the wrong trigonometric function, using $\sin(2\theta)$ instead of $\cos(2\theta)$. Study tip: Always remember the Jacobian! In polar coordinates, $dA$ becomes $r \, dr \, d\theta$, which adds an extra factor of $r$ to your integrand. This is the most commonly missed step in polar coordinate conversions.

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