| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the given double integral using polar coordinates, we need to convert the integral limits and the integrand to polar coordinates. The semicircle of diameter 1 with center at C(1/2, 0) above the X-axis can be described in polar coordinates as follows: The radius of the semicircle is 0.5, since the diameter is 1. The semicircle is centered at C(1/2, 0), which means the polar coordinates of the center are $(\rho, \theta) = (\frac{1}{2}, 0)$. The semicircle is above the X-axis, so the angle $\theta$ varies from 0 to $\pi$. The polar coordinates $(\rho, \theta)$ can be transformed to Cartesian coordinates $(x, y)$ using the formulas: $x = \rho \cos(\theta)$ $y = \rho \sin(\theta)$ The Jacobian of the transformation is $\rho$, so the differential element $dxdy$ becomes $\rho d\rho d\theta$. The given double integral is $\iint_S ydxdy$ with $y > 0$. Since $y = \rho \sin(\theta)$, the integrand becomes $\rho^2 \sin(\theta) d\rho d\theta$. The limits of integration for $\rho$ are from 0 to 0.5 (the radius of the semicircle), and the limits of integration for $\theta$ are from 0 to $\pi$. Thus, the double integral in polar coordinates becomes: $\int_0^\pi \int_0^{0.5} \rho^2 \sin(\theta) d\rho d\theta$. Let's solve the integral: $\int_0^\pi \int_0^{0.5} \rho^2 \sin(\theta) d\rho d\theta = \int_0^\pi \sin(\theta) d\theta \int_0^{0.5} \rho^2 d\rho = \left[ -\cos(\theta) \right]_0^\pi \left[ \frac{\rho^3}{3} \right]_0^{0.5} = (-\cos(\pi) + \cos(0)) \frac{(0.5)^3}{3} = 2 \cdot \frac{1}{24} = \frac{1}{12}$. Therefore, the value of the given double integral is $\boxed{\frac{1}{12}}$. |
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| deepseekr1 | To solve the double integral \(\iint_S y \, dx \, dy\) over the semicircle \(S\) with diameter 1, center at \((1/2, 0)\), and \(y > 0\) using polar coordinates, we proceed as follows: 1. **Equation of the Semicircle**: The semicircle is described by \((x - 1/2)^2 + y^2 = (1/2)^2\) with \(y \geq 0\). Converting this to polar coordinates \((r, \theta)\): \[ (r \cos \theta - 1/2)^2 + (r \sin \theta)^2 = (1/2)^2 \] Simplifying this equation, we get: \[ r = \cos \theta \] 2. **Limits of Integration**: - The radius \(r\) ranges from 0 to \(\cos \theta\). - The angle \(\theta\) ranges from 0 to \(\pi/2\) because the semicircle is above the x-axis and centered at \((1/2, 0)\). 3. **Integral Conversion**: - Convert \(y\) to polar coordinates: \(y = r \sin \theta\). - The Jacobian determinant for polar coordinates is \(r\), so \(dx \, dy = r \, dr \, d\theta\). 4. **Setting Up the Integral**: \[ \iint_S y \, dx \, dy = \int_{0}^{\pi/2} \int_{0}^{\cos \theta} (r \sin \theta) \cdot r \, dr \, d\theta \] Simplifying the integrand: \[ \int_{0}^{\pi/2} \int_{0}^{\cos \theta} r^2 \sin \theta \, dr \, d\theta \] 5. **Evaluating the Inner Integral**: \[ \int_{0}^{\cos \theta} r^2 \, dr = \left[ \frac{r^3}{3} \right]_{0}^{\cos \theta} = \frac{\cos^3 \theta}{3} \] 6. **Evaluating the Outer Integral**: \[ \int_{0}^{\pi/2} \frac{\cos^3 \theta}{3} \sin \theta \, d\theta \] Using substitution \(u = \cos \theta\), \(du = -\sin \theta \, d\theta\): \[ \frac{1}{3} \int_{1}^{0} u^3 (-du) = \frac{1}{3} \int_{0}^{1} u^3 \, du = \frac{1}{3} \left[ \frac{u^4}{4} \right]_{0}^{1} = \frac{1}{12} \] Thus, the value of the double integral is \(\boxed{\dfrac{1}{12}}\). |
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| gpto1 | To solve the integral \(\iint_S y\,dx\,dy\) over the semicircle \(S\), we'll convert the integral into polar coordinates. The semicircle \(S\) is defined as the upper half of the circle centered at \(C\left(\frac{1}{2}, 0\right)\) with radius \(\frac{1}{2}\), i.e., \[ (x - \tfrac{1}{2})^2 + y^2 = \left(\tfrac{1}{2}\right)^2, \quad y \geq 0. \] **Step 1: Shift Coordinates** To simplify, shift the coordinate system so that the circle is centered at the origin. Let: \[ u = x - \tfrac{1}{2}, \] so the circle equation becomes: \[ u^2 + y^2 = \left(\tfrac{1}{2}\right)^2. \] **Step 2: Convert to Polar Coordinates** In the shifted coordinates \((u, y)\), introduce polar coordinates: \[ u = r\cos\theta, \quad y = r\sin\theta, \] where \(r \geq 0\) and \(\theta\) ranges from \(0\) to \(\pi\) (upper semicircle). The Jacobian determinant for the change of variables is \(J = r\), so \(du\,dy = r\,dr\,d\theta\). **Step 3: Express the Integral in Polar Coordinates** The integral transforms to: \[ I = \iint_S y\,dx\,dy = \iint_{D} y\,du\,dy = \int_{0}^{\pi}\int_{0}^{\frac{1}{2}} (r\sin\theta) \cdot r\,dr\,d\theta = \int_{0}^{\pi}\int_{0}^{\frac{1}{2}} r^2\sin\theta\,dr\,d\theta. \] **Step 4: Compute the Integral** First, integrate with respect to \(r\): \[ \int_{0}^{\frac{1}{2}} r^2\,dr = \left[\frac{1}{3}r^3\right]_0^{\frac{1}{2}} = \frac{1}{24}. \] Then, integrate with respect to \(\theta\): \[ I = \frac{1}{24}\int_{0}^{\pi}\sin\theta\,d\theta = \frac{1}{24}\left[-\cos\theta\right]_0^{\pi} = \frac{1}{24}(-\cos\pi + \cos 0) = \frac{1}{24}(1 + 1) = \frac{1}{12}. \] **Final Answer:** \[ I = \frac{1}{12}. \] **Answer:** \(\boxed{\dfrac{1}{12}}\) |
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| GPT-4o | To solve the given problem of evaluating the double integral \(\iint_S y \, dx \, dy\) over the region \(S\), which is a semicircle with diameter 1 centered at \((\frac{1}{2}, 0)\) and lies above the x-axis, we will convert the integral to polar coordinates. ### Step 1: Describe the region \(S\) in Cartesian coordinates The semicircle has its center at \((\frac{1}{2}, 0)\) and radius \(\frac{1}{2}\). The equation of the full circle is: \[ (x - \frac{1}{2})^2 + y^2 = \left(\frac{1}{2}\right)^2 \] Simplifying, we get: \[ (x - \frac{1}{2})^2 + y^2 = \frac{1}{4} \] The region \(S\) is the upper half of this circle, so \(y \geq 0\). ### Step 2: Convert to polar coordinates To convert to polar coordinates, we shift the center of the semicircle to the origin. Let: \[ x' = x - \frac{1}{2} \] Then \(x'^2 + y^2 = \frac{1}{4}\). In polar coordinates \((r, \theta)\), where \(r\) is the radius and \(\theta\) is the angle, we have: \[ x' = r \cos \theta \quad \text{and} \quad y = r \sin \theta \] with \(0 \leq r \leq \frac{1}{2}\) and \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\) since the semicircle lies above the x-axis. The coordinates \(x\) and \(y\) in terms of \(r\) and \(\theta\) are: \[ x = x' + \frac{1}{2} = r \cos \theta + \frac{1}{2} \] \[ y = r \sin \theta \] ### Step 3: Jacobian determinant for polar coordinates The Jacobian determinant for polar coordinates is \(r\), so the area element \(dx \, dy\) becomes \(r \, dr \, d\theta\). ### Step 4: Set up the integral in polar coordinates The integral \(\iint_S y \, dx \, dy\) in polar coordinates is: \[ \iint_S y \, dx \, dy = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{\frac{1}{2}} (r \sin \theta) \cdot r \, dr \, d\theta \] ### Step 5: Evaluate the integral Separate the integral into two parts: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{\frac{1}{2}} r^2 \sin \theta \, dr \, d\theta \] First, integrate with respect to \(r\): \[ \int_{0}^{\frac{1}{2}} r^2 \, dr = \left[ \frac{r^3}{3} \right]_{0}^{\frac{1}{2}} = \frac{1}{3} \left(\frac{1}{2} \right)^3 = \frac{1}{3} \cdot \frac{1}{8} = \frac{1}{24} \] Next, integrate with respect to \(\theta\): \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin \theta \, d\theta \] The integral of \(\sin \theta\) over \([- \frac{\pi}{2}, \frac{\pi}{2}]\) is: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin \theta \, d\theta = \left[ - \cos \theta \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \left( -\cos \left(\frac{\pi}{2}\right) - \left( - \cos \left(- \frac{\pi}{2}\right) \right) \right) \] Since \(\cos(\frac{\pi}{2}) = 0\) and \(\cos(- \frac{\pi}{2}) = 0\): \[ 0 - 0 = 0 \] Re-evaluate within the correct limits: \[ \left[ - \cos \theta \right |
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| llama3.1-70B-Instruct | To solve the given double integral $\iint_S ydxdy$ where $S$ is a semicircle of diameter 1 with center at point $C(1/2, 0)$ above the X-axis, we first convert the problem to polar coordinates. The conversion formulas are $x = r\cos(\theta)$ and $y = r\sin(\theta)$. Given that the semicircle has a diameter of 1, its radius is $1/2$. The center of the semicircle is at $(1/2, 0)$, which means in polar coordinates, the center is at $(r, \theta) = (1/2, 0)$ for the point directly on the x-axis, but since we're dealing with a semicircle above the x-axis, we consider $\theta$ ranging from $0$ to $\pi$. However, because the semicircle's center is not at the origin, we need to adjust our approach. The integral in Cartesian coordinates is given, but converting directly to polar might be tricky due to the center's offset. Instead, we recognize that the integral of $y$ over the semicircle can be approached by considering the geometry of the problem. The integral $\iint_S ydxdy$ essentially calculates the volume under the plane $z=y$ over the semicircle $S$. However, converting this directly to polar coordinates without a clear origin at the center of the semicircle complicates the integration bounds. A simpler approach is to recognize that the integral of $y$ over the semicircle (which is symmetric about its vertical axis) can be calculated by considering the area of the semicircle and its average height. The average height of the semicircle above the x-axis is the radius of the semicircle, which is $1/2$. The area of the semicircle is $\frac{1}{2}\pi r^2 = \frac{1}{2}\pi(\frac{1}{2})^2 = \frac{\pi}{8}$. Thus, the integral $\iint_S ydxdy$ can be thought of as the area of the semicircle times its average height, which is $\frac{\pi}{8} \times \frac{1}{2} = \frac{\pi}{16}$. This approach bypasses the need for a direct conversion to polar coordinates and leverages the geometric properties of the semicircle to solve the problem. |
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| llama3.1-405B-Instruct | To solve this problem, we need to convert the given integral to polar coordinates and then evaluate it. First, let's convert the integral to polar coordinates. We have: $$\iint_S ydxdy = \int_{0}^{\pi/2} \int_{0}^{1/2} r \sin \theta \cdot r dr d\theta$$ where $r$ is the radial distance from the origin, $\theta$ is the angle from the positive x-axis, and $y = r \sin \theta$. Now, we need to find the limits of integration for $r$ and $\theta$. The semicircle has a diameter of 1, so the radius is 1/2. The center of the semicircle is at (1/2, 0), so the distance from the origin to the center is 1/2. Therefore, the limits of integration for $r$ are from 0 to 1/2. The limits of integration for $\theta$ are from 0 to $\pi/2$ because we are only considering the upper half of the semicircle. Now, we can evaluate the integral: $$\int_{0}^{\pi/2} \int_{0}^{1/2} r \sin \theta \cdot r dr d\theta = \int_{0}^{\pi/2} \left[ \frac{r^3}{3} \sin \theta \right]_{0}^{1/2} d\theta$$ $$= \int_{0}^{\pi/2} \frac{1}{24} \sin \theta d\theta = \left[ -\frac{1}{24} \cos \theta \right]_{0}^{\pi/2}$$ $$= -\frac{1}{24} \cos \frac{\pi}{2} + \frac{1}{24} \cos 0 = \frac{1}{24}$$ Therefore, the value of the double integral is $\frac{1}{24}$. |
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| DeepSeek-Coder-V2 | To solve the given problem, we need to calculate the double integral \(\iint_S y \, dx \, dy\) where \(S\) is a semicircle of diameter 1 with center at point \(C(1/2, 0)\) above the X axis. We will use polar coordinates to simplify the integration. ### Step 1: Describe the Region \(S\) The semicircle \(S\) has a diameter of 1 and is centered at \(C(1/2, 0)\). The radius of the semicircle is \(1/2\). The equation of the circle in Cartesian coordinates is: \[ (x - 1/2)^2 + y^2 = (1/2)^2 \] Since \(y > 0\), we are only considering the upper half of this circle. ### Step 2: Convert to Polar Coordinates In polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\). The center of the circle is at \((1/2, 0)\), so we need to shift the origin to this point. The new coordinates are: \[ x' = x - 1/2 \quad \text{and} \quad y' = y \] Thus, the equation of the circle becomes: \[ (x')^2 + y'^2 = (1/2)^2 \] In polar coordinates, this is: \[ (r \cos \theta)^2 + (r \sin \theta)^2 = (1/2)^2 \] \[ r^2 (\cos^2 \theta + \sin^2 \theta) = (1/2)^2 \] \[ r^2 = (1/2)^2 \] \[ r = 1/2 \] ### Step 3: Determine the Limits of Integration The angle \(\theta\) ranges from \(0\) to \(\pi\) because we are considering the semicircle above the X axis. The radius \(r\) ranges from \(0\) to \(1/2\). ### Step 4: Set Up the Integral The double integral in polar coordinates is: \[ \iint_S y \, dx \, dy = \int_0^\pi \int_0^{1/2} (r \sin \theta) r \, dr \, d\theta \] ### Step 5: Evaluate the Integral First, integrate with respect to \(r\): \[ \int_0^{1/2} r^2 \sin \theta \, dr \] \[ = \sin \theta \int_0^{1/2} r^2 \, dr \] \[ = \sin \theta \left[ \frac{r^3}{3} \right]_0^{1/2} \] \[ = \sin \theta \left( \frac{(1/2)^3}{3} \right) \] \[ = \sin \theta \left( \frac{1/8}{3} \right) \] \[ = \frac{1}{24} \sin \theta \] Next, integrate with respect to \(\theta\): \[ \int_0^\pi \frac{1}{24} \sin \theta \, d\theta \] \[ = \frac{1}{24} \int_0^\pi \sin \theta \, d\theta \] \[ = \frac{1}{24} \left[ -\cos \theta \right]_0^\pi \] \[ = \frac{1}{24} \left( -\cos \pi - (-\cos 0) \right) \] \[ = \frac{1}{24} \left( -(-1) - (-1) \right) \] \[ = \frac{1}{24} \left( 1 + 1 \right) \] \[ = \frac{1}{24} \cdot 2 \] \[ = \frac{1}{12} \] Thus, the value of the double integral is: \[ \boxed{\frac{1}{12}} \] |
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| Qwen2-72B-Instruct | To solve the given double integral using polar coordinates, we need to convert the integral limits and the integrand to polar coordinates. The semicircle of diameter 1 with center at C(1/2, 0) above the X-axis can be described in polar coordinates as follows: The radius of the semicircle is 0.5, since the diameter is 1. The semicircle is centered at C(1/2, 0), which means the polar coordinates of the center are $(\rho, \theta) = (\frac{1}{2}, 0)$. The semicircle is above the X-axis, so the angle $\theta$ varies from 0 to $\pi$. The polar coordinates $(\rho, \theta)$ can be transformed to Cartesian coordinates $(x, y)$ using the formulas: $x = \rho \cos(\theta)$ $y = \rho \sin(\theta)$ The Jacobian of the transformation is $\rho$, so the differential element $dxdy$ becomes $\rho d\rho d\theta$. The given double integral is $\iint_S ydxdy$ with $y > 0$. Since $y = \rho \sin(\theta)$, the integrand becomes $\rho^2 \sin(\theta) d\rho d\theta$. The limits of integration for $\rho$ are from 0 to 0.5 (the radius of the semicircle), and the limits of integration for $\theta$ are from 0 to $\pi$. Thus, the double integral in polar coordinates becomes: $\int_0^\pi \int_0^{0.5} \rho^2 \sin(\theta) d\rho d\theta$. Let's solve the integral: $\int_0^\pi \int_0^{0.5} \rho^2 \sin(\theta) d\rho d\theta = \int_0^\pi \sin(\theta) d\theta \int_0^{0.5} \rho^2 d\rho = \left[ -\cos(\theta) \right]_0^\pi \left[ \frac{\rho^3}{3} \right]_0^{0.5} = (-\cos(\pi) + \cos(0)) \frac{(0.5)^3}{3} = 2 \cdot \frac{1}{24} = \frac{1}{12}$. Therefore, the value of the given double integral is $\boxed{\frac{1}{12}}$. |
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| baichuan4 | To solve the given problem, we will convert the integral to polar coordinates. The semicircle S has its center at point C(1/2, 0) and a diameter of 1, which means its radius r is 1/2. The semicircle is above the X axis, indicating that the angle θ ranges from 0 to π. In polar coordinates, the integral becomes: $$ \iint_S y dx dy = \int_0^\pi \int_0^{1/2} r^2 \sin(\theta) dr d\theta $$ Now, we can calculate the integral: $$ \int_0^\pi \int_0^{1/2} r^2 \sin(\theta) dr d\theta = \int_0^\pi \left[ \frac{r^3}{3} \right]_0^{1/2} \sin(\theta) d\theta $$ $$ = \int_0^\pi \frac{1}{24} \sin(\theta) d\theta $$ $$ = \left[ -\frac{1}{24} \cos(\theta) \right]_0^\pi $$ $$ = -\frac{1}{24} (-1 - 1) $$ $$ = \frac{1}{12} $$ So, the value of the double integral is 1/12. |
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| doubao | The equation of the semicircle in Cartesian coordinates is $(x - \frac{1}{2})^2 + y^2 = (\frac{1}{2})^2$. Converting to polar coordinates, $x = \frac{1}{2} + r\cos\theta$ and $y = r\sin\theta$. The semicircle has radius $\frac{1}{2}$. The limits for $r$ are from 0 to $\frac{1}{2}$ and for $\theta$ are from 0 to $\pi$. The integral becomes: \[ \begin{align*} &\int_{0}^{\pi}\int_{0}^{\frac{1}{2}}r\sin\theta \cdot r dr d\theta\\ =&\int_{0}^{\pi}\sin\theta \left[\frac{r^3}{3}\right]_{0}^{\frac{1}{2}} d\theta\\ =&\int_{0}^{\pi}\sin\theta \left(\frac{(\frac{1}{2})^3}{3}\right) d\theta\\ =&\frac{1}{24}\int_{0}^{\pi}\sin\theta d\theta\\ =&\frac{1}{24} [-\cos\theta]_{0}^{\pi}\\ =&\frac{1}{24} (1 - (-1))\\ =&\frac{1}{12} \end{align*} \] |
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| minimax | To solve the double integral $\iint_S y \, dx \, dy$ over the semicircle $S$ with a diameter of 1 and centered at $C(1/2, 0)$ above the x-axis, we first need to define the region $S$ in polar coordinates. Since the semicircle is centered at $(1/2, 0)$, the radius $r$ will range from $0$ to $1/2$, and the angle $\theta$ will range from $0$ to $\pi$ to cover the upper semicircle. In polar coordinates, the relationship between Cartesian coordinates $(x, y)$ and polar coordinates $(r, \theta)$ is given by: $$ x = r \cos(\theta) $$ $$ y = r \sin(\theta) $$ The Jacobian determinant for the transformation from Cartesian to polar coordinates is $r$, so the differential area element $dx \, dy$ becomes $r \, dr \, d\theta$ in polar coordinates. The integral becomes: $$ \iint_S y \, dx \, dy = \int_{0}^{\pi} \int_{0}^{1/2} (r \sin(\theta)) \, r \, dr \, d\theta $$ Now we can compute the integral: $$ \int_{0}^{\pi} \sin(\theta) \, d\theta \int_{0}^{1/2} r^2 \, dr $$ First, integrate with respect to $r$: $$ \int_{0}^{1/2} r^2 \, dr = \left[ \frac{r^3}{3} \right]_{0}^{1/2} = \frac{(1/2)^3}{3} - \frac{0^3}{3} = \frac{1}{24} $$ Now, integrate with respect to $\theta$: $$ \int_{0}^{\pi} \sin(\theta) \, d\theta = \left[ -\cos(\theta) \right]_{0}^{\pi} = -\cos(\pi) + \cos(0) = 1 + 1 = 2 $$ Multiplying the two results together gives the value of the integral: $$ \iint_S y \, dx \, dy = \frac{1}{24} \cdot 2 = \frac{1}{12} $$ Therefore, the value of the double integral over the given semicircle region $S$ is $\frac{1}{12}$. |
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| kimi | To solve the double integral $\iint_S y \, dx \, dy$ over the region \( S \), which is a semicircle of diameter 1 centered at point \( C(\frac{1}{2}, 0) \) above the \( x \)-axis, we first need to express the region \( S \) in polar coordinates. In polar coordinates, a point \((x, y)\) is represented as \((r, \theta)\), where \( r \) is the distance from the origin and \( \theta \) is the angle from the positive \( x \)-axis. The region \( S \) can be described as follows: 1. \( r \) ranges from 0 to \( \sqrt{(1/2 - x)^2 + y^2} \), since the radius is the distance from the center of the semicircle to any point on the semicircle. 2. \( \theta \) ranges from \( \pi \) to \( 2\pi \) because the semicircle is above the \( x \)-axis. However, since the center of the semicircle is not at the origin but at \( (\frac{1}{2}, 0) \), we need to shift our polar coordinates accordingly. The new center \( C \) is at \( (\frac{1}{2}, 0) \), so we need to translate the polar coordinates by \( \frac{1}{2} \) units along the \( x \)-axis. The equation of the semicircle in Cartesian coordinates is: \[ (x - \frac{1}{2})^2 + y^2 = \left(\frac{1}{2}\right)^2 \] \[ y^2 = \left(\frac{1}{2}\right)^2 - (x - \frac{1}{2})^2 \] \[ y = \pm \sqrt{\left(\frac{1}{2}\right)^2 - (x - \frac{1}{2})^2} \] Since \( y > 0 \), we only consider the positive root. Now, we convert this to polar coordinates by shifting the origin to \( C \). The new polar coordinates \( (r', \theta') \) are related to the original coordinates \( (r, \theta) \) by: \[ r' = \sqrt{(x - \frac{1}{2})^2 + y^2} \] \[ \theta' = \theta - \pi \] The integral in polar coordinates becomes: \[ \iint_S y \, dx \, dy = \int_{\pi}^{2\pi} \int_0^{\sqrt{(1/2)^2 - (1/2 - r\cos(\theta))^2}} r' \cdot r' \cos(\theta') \, dr' \, d\theta' \] Since \( y = r' \cos(\theta') \) and \( dx \, dy = r' \, dr' \, d\theta' \), the integral simplifies to: \[ \int_{\pi}^{2\pi} \int_0^{\sqrt{(1/2)^2 - (1/2 - r\cos(\theta))^2}} r'^2 \cos(\theta') \, dr' \, d\theta' \] Now we can evaluate the integral. The inner integral is with respect to \( r' \), and the outer integral is with respect to \( \theta' \). However, it's important to note that the limits of integration for \( r' \) are not straightforward due to the square root and the fact that the center is not at the origin. We need to find the correct limits for \( r' \) by solving the equation of the semicircle for \( r' \) when \( \theta' \) is given. The correct limits for \( r' \) are from 0 to \( \frac{1}{2} \) because the radius of the semicircle is \( \frac{1}{2} \). So the integral becomes: \[ \int_{\pi}^{2\pi} \int_0^{1/2} \frac{1}{2} r^2 \cos(\theta) \, dr \, d\theta \] Now we can evaluate the integral: \[ \int_{\pi}^{2\pi} \left[ \frac{1}{6} r^3 \right]_0^{1/2} \cos(\theta) \, d\theta \] \[ = \int_{\pi}^{2\pi} \frac{1}{6 |
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| wenxinyiyan | Let's solve the given problem by passing to polar coordinates. First, we need to express the semicircle S in polar coordinates. The semicircle has a diameter of 1 and is centered at C(1/2, 0). In polar coordinates, we can represent this as: $r = \frac{1}{2}\cos\theta + \frac{\sqrt{1 - (\frac{1}{2}\sin\theta)^2}}{2}$ However, since the semicircle is above the X-axis and has a diameter of 1, a simpler representation is: $r = \frac{1}{2}$ with $\theta$ ranging from 0 to $\pi$ (because it's a semicircle above the X-axis). Now, let's convert the given integral to polar coordinates. The integral is: $\iint_S y\,dx\,dy$ In polar coordinates, $y = r\sin\theta$ and $dx\,dy = r\,dr\,d\theta$. So, the integral becomes: $\int_0^\pi \int_0^{1/2} r\sin\theta \,r\,dr\,d\theta$ Let's solve this integral step by step: 1. First, integrate with respect to $r$: $\int_0^{1/2} r^2\sin\theta \,dr = \left[\frac{r^3}{3}\sin\theta\right]_0^{1/2} = \frac{1}{24}\sin\theta$ 2. Now, integrate with respect to $\theta$: $\int_0^\pi \frac{1}{24}\sin\theta \,d\theta = \left[-\frac{1}{24}\cos\theta\right]_0^\pi = \frac{1}{12}$ So, the value of the double integral $\iint_S y\,dx\,dy$ with $y > 0$ over the semicircle S is $\frac{1}{12}$. |
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| zhipu | To solve the double integral \(\iint_S y \, dx \, dy\) over the semicircle \(S\) with a diameter of 1 and center at \(C\left(\frac{1}{2}, 0\right)\), we can convert the integral to polar coordinates. Here's a step-by-step solution: ### Step 1: Describe the Region \(S\) in Cartesian Coordinates The semicircle \(S\) has a diameter of 1, so its radius \(r\) is \(\frac{1}{2}\). The center of the semicircle is at \(\left(\frac{1}{2}, 0\right)\). The equation of the semicircle in Cartesian coordinates is: \[ \left(x - \frac{1}{2}\right)^2 + y^2 = \left(\frac{1}{2}\right)^2 \] This simplifies to: \[ \left(x - \frac{1}{2}\right)^2 + y^2 = \frac{1}{4} \] ### Step 2: Convert to Polar Coordinates In polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\). The equation of the semicircle becomes: \[ \left(r \cos \theta - \frac{1}{2}\right)^2 + (r \sin \theta)^2 = \frac{1}{4} \] Expanding and simplifying: \[ r^2 \cos^2 \theta - r \cos \theta + \frac{1}{4} + r^2 \sin^2 \theta = \frac{1}{4} \] \[ r^2 (\cos^2 \theta + \sin^2 \theta) - r \cos \theta = 0 \] \[ r^2 - r \cos \theta = 0 \] \[ r (r - \cos \theta) = 0 \] Since \(r \neq 0\), we have: \[ r = \cos \theta \] ### Step 3: Determine the Limits of Integration For the semicircle above the x-axis, \(\theta\) ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). The radius \(r\) ranges from 0 to \(\cos \theta\). ### Step 4: Set Up the Integral in Polar Coordinates The integral in polar coordinates is: \[ \iint_S y \, dx \, dy = \iint_S r \sin \theta \cdot r \, dr \, d\theta \] \[ = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{\cos \theta} r^2 \sin \theta \, dr \, d\theta \] ### Step 5: Evaluate the Inner Integral First, integrate with respect to \(r\): \[ \int_0^{\cos \theta} r^2 \sin \theta \, dr = \sin \theta \int_0^{\cos \theta} r^2 \, dr \] \[ = \sin \theta \left[ \frac{r^3}{3} \right]_0^{\cos \theta} \] \[ = \sin \theta \cdot \frac{(\cos \theta)^3}{3} \] \[ = \frac{\sin \theta \cos^3 \theta}{3} \] ### Step 6: Evaluate the Outer Integral Now, integrate with respect to \(\theta\): \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin \theta \cos^3 \theta}{3} \, d\theta \] Factor out the constant \(\frac{1}{3}\): \[ = \frac{1}{3} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin \theta \cos^3 \theta \, d\theta \] Use the substitution \(u = \cos \theta\), \(du = -\sin \theta \, d\theta\): \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin \theta \cos^3 \theta \, d\theta = \int_{\cos(-\frac{\pi}{2})}^{\cos(\frac{\pi}{2})} u^3 (-du) \] \[ = \int_0^1 u^3 \, du \] \[ = \left[ \frac{u^4}{4} \right]_0^1 \] \[ = \frac{1}{4} \] ### Step 7: Combine the Results \[ \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12} \] ### Final Answer The value of the double integral \(\iint_S y \, dx \, dy\) is: \[ \boxed{\frac{1}{12}} \] |
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| gpto1 | To solve the integral \(\iint_S y\,dx\,dy\) over the semicircle \(S\), we'll convert the integral into polar coordinates. The semicircle \(S\) is defined as the upper half of the circle centered at \(C\left(\frac{1}{2}, 0\right)\) with radius \(\frac{1}{2}\), i.e., \[ (x - \tfrac{1}{2})^2 + y^2 = \left(\tfrac{1}{2}\right)^2, \quad y \geq 0. \] **Step 1: Shift Coordinates** To simplify, shift the coordinate system so that the circle is centered at the origin. Let: \[ u = x - \tfrac{1}{2}, \] so the circle equation becomes: \[ u^2 + y^2 = \left(\tfrac{1}{2}\right)^2. \] **Step 2: Convert to Polar Coordinates** In the shifted coordinates \((u, y)\), introduce polar coordinates: \[ u = r\cos\theta, \quad y = r\sin\theta, \] where \(r \geq 0\) and \(\theta\) ranges from \(0\) to \(\pi\) (upper semicircle). The Jacobian determinant for the change of variables is \(J = r\), so \(du\,dy = r\,dr\,d\theta\). **Step 3: Express the Integral in Polar Coordinates** The integral transforms to: \[ I = \iint_S y\,dx\,dy = \iint_{D} y\,du\,dy = \int_{0}^{\pi}\int_{0}^{\frac{1}{2}} (r\sin\theta) \cdot r\,dr\,d\theta = \int_{0}^{\pi}\int_{0}^{\frac{1}{2}} r^2\sin\theta\,dr\,d\theta. \] **Step 4: Compute the Integral** First, integrate with respect to \(r\): \[ \int_{0}^{\frac{1}{2}} r^2\,dr = \left[\frac{1}{3}r^3\right]_0^{\frac{1}{2}} = \frac{1}{24}. \] Then, integrate with respect to \(\theta\): \[ I = \frac{1}{24}\int_{0}^{\pi}\sin\theta\,d\theta = \frac{1}{24}\left[-\cos\theta\right]_0^{\pi} = \frac{1}{24}(-\cos\pi + \cos 0) = \frac{1}{24}(1 + 1) = \frac{1}{12}. \] **Final Answer:** \[ I = \frac{1}{12}. \] **Answer:** \(\boxed{\dfrac{1}{12}}\) |
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