вычислить $\int_L (x+y) \, dL$ , где $L$ - правый лепесток $r^2 = a^2\cos(2\varphi)$

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Find derivatives using the quotient rule I

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Q. вычислить

\int_L (x+y) \, dL

, где

L

- правый лепесток

r^2 = a^2\cos(2\varphi)

Convert to Polar Coordinates: Let's first express the integral in polar coordinates since the given curve is already in polar form. In polar coordinates, $x = r\cos(\varphi)$ and $y = r\sin(\varphi)$ . The curve $L$ is defined by $r^2 = a^2\cos(2\varphi)$ , which simplifies to $r = a\sqrt{\cos(2\varphi)}$ for the right petal. We need to find the bounds for $\varphi$ that correspond to the right petal.
Find Bounds for $\varphi$ : The right petal of the curve $r^2 = a^2\cos(2\varphi)$ corresponds to the values of $\varphi$ for which $\cos(2\varphi)$ is positive. This occurs when $2\varphi$ is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ , which means $\varphi$ is between $-\frac{\pi}{4}$ and $\frac{\pi}{4}$ . These are the bounds for our integral.
Compute $\frac{dr}{d\varphi}$ : Now we can set up the integral in polar coordinates. The differential arc length in polar coordinates is given by $dL = \sqrt{r^2 + \left(\frac{dr}{d\varphi}\right)^2} d\varphi$ . We need to compute $\frac{dr}{d\varphi}$ for our curve.
Simplify Expression for $dL$ : Differentiating $r$ with respect to $\varphi$ , we get $\frac{dr}{d\varphi} = \frac{d}{d\varphi} (a\sqrt{\cos(2\varphi)}) = a\left(-\sin(2\varphi)\right)/\sqrt{\cos(2\varphi)}$ . We need to square this and add it to $r^2$ to find the expression for $dL$ .
Write Integral in Polar Coordinates: Squaring $\frac{dr}{d\varphi}$ , we get $\left(\frac{dr}{d\varphi}\right)^2 = a^2\sin^2(2\varphi)/\cos(2\varphi)$ . Adding this to $r^2$ , we get $r^2 + \left(\frac{dr}{d\varphi}\right)^2 = a^2\cos(2\varphi) + a^2\sin^2(2\varphi)/\cos(2\varphi) = a^2\cos(2\varphi) + a^2\tan^2(2\varphi)$ .
Substitute $r$ and Simplify: We can simplify the expression for $dL$ using the identity $1 + \tan^2(\varphi) = \sec^2(\varphi)$ . So, $r^2 + (\frac{dr}{d\varphi})^2 = a^2\cos(2\varphi) + a^2\tan^2(2\varphi) = a^2\sec^2(2\varphi)$ . Therefore, $dL = a\sqrt{\sec^2(2\varphi)} d\varphi$ .
Integrate the Expression: Now we can write the integral of $(x+y)$ along the curve $L$ in polar coordinates as $\int (r\cos(\varphi) + r\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ .
Evaluate the Integral: Substituting $r = a\sqrt{\cos(2\varphi)}$ , the integral becomes $\int_{\varphi = -\frac{\pi}{4}}^{\varphi = \frac{\pi}{4}} \left(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi)\right)\sqrt{\sec^2(2\varphi)} \, d\varphi$ .
Evaluate the Integral: Substituting $r = a\sqrt{\cos(2\varphi)}$ , the integral becomes $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ .We can simplify the integral by noting that $\sqrt{\sec^2(2\varphi)} = |\sec(2\varphi)|$ . Since we are integrating over the range where $\cos(2\varphi)$ is positive, we can drop the absolute value and write $\sqrt{\sec^2(2\varphi)}$ as $\sec(2\varphi)$ .
Evaluate the Integral: Substituting $r = a\sqrt{\cos(2\varphi)}$ , the integral becomes $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} d\varphi$ from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ .We can simplify the integral by noting that $\sqrt{\sec^2(2\varphi)} = |\sec(2\varphi)|$ . Since we are integrating over the range where $\cos(2\varphi)$ is positive, we can drop the absolute value and write $\sqrt{\sec^2(2\varphi)}$ as $\sec(2\varphi)$ .The integral simplifies to $\int a\sqrt{\cos(2\varphi)}(\cos(\varphi) + \sin(\varphi))\sec(2\varphi) d\varphi$ from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ . We can now integrate this expression with respect to $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} d\varphi$ $1$ .
Evaluate the Integral: Substituting $r = a\sqrt{\cos(2\varphi)}$ , the integral becomes $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ .We can simplify the integral by noting that $\sqrt{\sec^2(2\varphi)} = |\sec(2\varphi)|$ . Since we are integrating over the range where $\cos(2\varphi)$ is positive, we can drop the absolute value and write $\sqrt{\sec^2(2\varphi)}$ as $\sec(2\varphi)$ .The integral simplifies to $\int a\sqrt{\cos(2\varphi)}(\cos(\varphi) + \sin(\varphi))\sec(2\varphi) \, d\varphi$ from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ . We can now integrate this expression with respect to $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ $1$ .The integral $\int a\sqrt{\cos(2\varphi)}(\cos(\varphi) + \sin(\varphi))\sec(2\varphi) \, d\varphi$ is not straightforward to integrate directly. We may need to use a substitution or simplify the integrand further. However, we notice that $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ $3$ , so the integral simplifies to $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ $4$ .
Evaluate the Integral: Substituting $r = a\sqrt{\cos(2\varphi)}$ , the integral becomes $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ .We can simplify the integral by noting that $\sqrt{\sec^2(2\varphi)} = |\sec(2\varphi)|$ . Since we are integrating over the range where $\cos(2\varphi)$ is positive, we can drop the absolute value and write $\sqrt{\sec^2(2\varphi)}$ as $\sec(2\varphi)$ .The integral simplifies to $\int a\sqrt{\cos(2\varphi)}(\cos(\varphi) + \sin(\varphi))\sec(2\varphi) \, d\varphi$ from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ . We can now integrate this expression with respect to $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ $1$ .The integral $\int a\sqrt{\cos(2\varphi)}(\cos(\varphi) + \sin(\varphi))\sec(2\varphi) \, d\varphi$ is not straightforward to integrate directly. We may need to use a substitution or simplify the integrand further. However, we notice that $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ $3$ , so the integral simplifies to $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ $4$ .Integrating $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ $4$ from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ , we get $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ $8$ evaluated from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ .
Evaluate the Integral: Substituting $r = a\sqrt{\cos(2\varphi)}$ , the integral becomes $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ .We can simplify the integral by noting that $\sqrt{\sec^2(2\varphi)} = |\sec(2\varphi)|$ . Since we are integrating over the range where $\cos(2\varphi)$ is positive, we can drop the absolute value and write $\sqrt{\sec^2(2\varphi)}$ as $\sec(2\varphi)$ .The integral simplifies to $\int a\sqrt{\cos(2\varphi)}(\cos(\varphi) + \sin(\varphi))\sec(2\varphi) \, d\varphi$ from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ . We can now integrate this expression with respect to $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ $1$ .The integral $\int a\sqrt{\cos(2\varphi)}(\cos(\varphi) + \sin(\varphi))\sec(2\varphi) \, d\varphi$ is not straightforward to integrate directly. We may need to use a substitution or simplify the integrand further. However, we notice that $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ $3$ , so the integral simplifies to $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ $4$ .Integrating $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ $4$ from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ , we get $\int(a\sqrt{\cos(2\varphi)}\cos(\varphi) + a\sqrt{\cos(2\varphi)}\sin(\varphi))\sqrt{\sec^2(2\varphi)} \, d\varphi$ $8$ evaluated from $\varphi = -\frac{\pi}{4}$ to $\varphi = \frac{\pi}{4}$ .Evaluating the antiderivative at the bounds, we get $\varphi = -\frac{\pi}{4}$ $1$ .

вычислить ∫L(x+y) dL \int_L (x+y) \, dL ∫L​(x+y)dL, где L L L - правый лепесток r2=a2cos⁡(2φ) r^2 = a^2\cos(2\varphi) r2=a2cos(2φ)

вычислить $\int_L (x+y) \, dL$ , где $L$ - правый лепесток $r^2 = a^2\cos(2\varphi)$