If C is the part of the circle (x/5)^2 + (y/5)^2 = 1 in the first quadrant, find the following line integral with respect to arc length. integral_c (8x - 3y)ds = _______.

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Question:

If C is the part of the circle (x/5)^2 + (y/5)^2 = 1 in the first quadrant, find the following line integral with respect to arc length. integral_c (8x - 3y)ds = _______.

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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]

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