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Calculus Level 4

Let C C be the anti clockwise semicircle represented by x 2 + y 2 = 1 , y 0 x^2 + y^2 = 1, y \geq 0 . If d S dS is the differential of the arc length of the curve, what is the value of

N = C ( 2 + x 2 y ) d S ? N = \int_C (2+x^2y)\ dS ?


The answer is 6.94985197.

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1 solution

Tunk-Fey Ariawan
Jan 31, 2014

C C is determined by parametric equations x = cos t x=\cos t , y = sin t y=\sin t , π 2 t π 2 -\frac{\pi}{2} \le t \le \frac{\pi}{2} , and d S = ( d x d t ) 2 + ( d y d t ) 2 d t dS=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt , then we obtain N = π 2 π 2 ( 2 + cos 2 t sin t ) ( sin t ) 2 + ( cos t ) 2 d t = 2 ( 0 π 2 2 d t 0 π 2 cos 2 t d ( cos t ) ) = 2 ( π 1 3 cos 3 t 0 π 2 ) = 2 ( π + 1 3 ) 6.95 \begin{aligned} N &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (2+ \cos^2 t \sin t)\sqrt{(-\sin t)^2+(\cos t)^2}\,dt\\ &=2\left(\int_{0}^{\frac{\pi}{2}}2\,dt-\int_{0}^{\frac{\pi}{2}}\cos^2 t\,d(\cos t)\right)\\ &=2\left(\pi-\left.\frac{1}{3}\cos^3 t\right|_0^{\frac{\pi}{2}}\right)\\ &=2\left(\pi+\frac{1}{3}\right)\\ &\approx \boxed{6.95} \end{aligned} # Q . E . D . # \text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}

i have also calculated the right answer but i confused what value of pi i should take

suraj upadhyay - 7 years, 3 months ago

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