If C is the part of the circle (x4)2+(y4)2=1 in the first quadrant, find the following line integral with respect to arc length. ∫C(8x−7y)ds=

Answer 1

Start with a parametric form for the path of integration; the standard one for the unit circle is:

x = cos(t) y = sin(t)

Now compute ds = sqrt(dx^2 + dy^2):

dx = -sin(t) dy = cos(t)

ds = sqrt(sin^2(t) + cos^2(t))dt = sqrt(1)dt = dt

The integrand is thus:

[8cos(t) - 7sin(t)]dt

An antiderivative is:

F(t) = 8sin(t) + 7cos(t)

In the first quadrant, 0 <= t <= pi/2. Assuming C goes anticlockwise from (1,0) to (0,1), the limits of integration are 0 (lower) to pi/2 (upper). The value of the integral is thus:

F(pi/2) - F(0) = [8(1) + 7(0)] - [8(0) + 7(1)] = 1

If C goes clockwise, swap the limits of integration (which negates the value of the integral).