Respuesta :
Convert to polar coordinates, in which the circle's equation becomes
[tex]\left(\dfrac x5\right)^2+\left(\dfrac y5\right)^2=1\implies x^2+y^2=5^2\implies r^2=5^2\implies r=5[/tex]
where [tex]x=5\cos\theta[/tex] and [tex]y=5\sin\theta[/tex], and we get the part of the circle in the first quadrant with [tex]0\le \theta\le\frac\pi2[/tex].
So the integral is
[tex]\displaystyle\int_C(8x-3y)\,\mathrm ds=\int_0^{\frac\pi2}(8x(\theta)-3y(\theta))\sqrt{\left(\dfrac{\mathrm dx}{\mathrm d\theta}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm d\theta}\right)^2}\,\mathrm d\theta[/tex]
[tex]=\displaystyle\int_0^{\frac\pi2}(40\cos\theta-15\sin\theta)\sqrt{25\cos^2\theta+25\sin^2\theta}\,\mathrm d\theta[/tex]
[tex]=\displaystyle25\int_0^{\frac\pi2}(8\cos\theta-3\sin\theta)\,\mathrm d\theta[/tex]
[tex]=25(8\sin\theta+3\cos\theta)\bigg|_0^{\frac\pi2}=200-75=\boxed{125}[/tex]
Line integral involves integrating a function along a curve
The value of the line integral is 125
The equation is given as:
[tex]\mathbf{(\frac{x}{5})^2 + (\frac{y}{5})^2 = 1}[/tex]
Expand
[tex]\mathbf{\frac{x^2}{5^2} + \frac{y^2}{5^2} = 1}[/tex]
Multiply through by 5^2
[tex]\mathbf{x^2 + y^2 = 5^2}[/tex]
The equation of a circle is represented as:
[tex]\mathbf{(x - a)^2 + (y - b)^2 = r^2}[/tex]
So, by comparison
[tex]\mathbf{r^2 = 5^2}[/tex]
[tex]\mathbf{r = 5}[/tex]
Where:
[tex]\mathbf{x = rcos\theta}[/tex]
[tex]\mathbf{y = rsin\theta}[/tex]
So, we have:
[tex]\mathbf{\int_c (8x - 3y)ds}[/tex]
Because it is in the first quadrant (i.e. 0 to pi/2), the integrand becomes
[tex]\mathbf{\int_c (8x - 3y)ds = \int\limits^{\frac{\pi}{2}}_0 (8x - 3y)rd\theta}[/tex]
Convert to polar forms
[tex]\mathbf{\int_c (8x - 3y)ds = \int\limits^{\frac{\pi}{2}}_0 (8x - 3y)\sqrt{x^2 + y^2}d\theta}[/tex]
Substitute [tex]\mathbf{x = rcos\theta}[/tex] and [tex]\mathbf{y = rsin\theta}[/tex]
[tex]\mathbf{\int_c (8x - 3y)ds = \int\limits^{\frac{\pi}{2}}_c (8rcos(\theta) - 3rsin(\theta))\sqrt{(rcos(\theta))^2 + ( rsin(\theta))^2}d\theta}[/tex]
Substitute 5 for r
[tex]\mathbf{\int_c (8x - 3y)ds = \int\limits^{\frac{\pi}{2}}_c (40cos(\theta) - 15sin(\theta))\sqrt{25cos^2\theta + 25sin^2\theta}\ d\theta}[/tex]
Factor out 5
[tex]\mathbf{\int_c (8x - 3y)ds = \int\limits^{\frac{\pi}{2}}_c 5(8cos(\theta) - 3sin(\theta))\sqrt{25cos^2\theta + 25sin^2\theta}\ d\theta}[/tex]
[tex]\mathbf{\int_c (8x - 3y)ds = \int\limits^{\frac{\pi}{2}}_c 5(8cos(\theta) - 3sin(\theta))\sqrt{25(cos^2\theta + sin^2\theta)}\ d\theta}[/tex]
[tex]\mathbf{\int_c (8x - 3y)ds = \int\limits^{\frac{\pi}{2}}_c 5(8cos(\theta) - 3sin(\theta))5\sqrt{(cos^2\theta + sin^2\theta)}\ d\theta}[/tex]
[tex]\mathbf{\int_c (8x - 3y)ds = \int\limits^{\frac{\pi}{2}}_c 25(8cos(\theta) - 3sin(\theta))\sqrt{(cos^2\theta + sin^2\theta)}\ d\theta}[/tex]
[tex]\mathbf{\int_c (8x - 3y)ds =25 \int\limits^{\frac{\pi}{2}}_c (8cos(\theta) - 3sin(\theta))\sqrt{(cos^2\theta + sin^2\theta)}\ d\theta}[/tex]
In trigonometry
[tex]\mathbf{cos^2\theta + sin^2\theta = 1}[/tex]
So, we have:
[tex]\mathbf{\int_c (8x - 3y)ds =25 \int\limits^{\frac{\pi}{2}}_c (8cos(\theta) - 3sin(\theta))\sqrt{1}\ d\theta}[/tex]
[tex]\mathbf{\int_c (8x - 3y)ds =25 \int\limits^{\frac{\pi}{2}}_0 (8cos(\theta) - 3sin(\theta))\ d\theta}[/tex]
Integrate
[tex]\mathbf{\int_c (8x - 3y)ds =25 \times [ (8sin(\theta) + 3cos(\theta))\ } ]|\limits^{\frac{\pi}{2}}_0}[/tex]
Expand
[tex]\mathbf{\int_c (8x - 3y)ds =25 \times ([ 8sin(\frac{\pi}{2}) + 3cos(\frac{\pi}{2})] - ([ (8sin(0) + 3cos(0)])}[/tex]
[tex]\mathbf{\int_c (8x - 3y)ds =25 \times ([ 8\times 1 + 3\times 0)] - ([ (8\times 0 + 3\times 1])}[/tex]
[tex]\mathbf{\int_c (8x - 3y)ds =25 \times ([ 8] - [ 3])}[/tex]
[tex]\mathbf{\int_c (8x - 3y)ds =25 \times 5}[/tex]
[tex]\mathbf{\int_c (8x - 3y)ds =125}[/tex]
Hence, the value of the line integral is 125
Read more about line integrals at:
https://brainly.com/question/16571667
