Question

Evaluate \(\int_{C}\left(x^{2}+y^{2}\right) d s\) \(C:\) counterclockwise around the circle \(x^{2}+y^{2}=1\) from (1,0) to (0,1)

Short answer

Therefore, the value of the line integral \(\int_{C}\left(x^{2}+y^{2}\right) d s\) around the given path is \(\pi/2\).

Step-by-step solution

Parameterization of the Line Path

The line path C is the unit circle \(x^2 + y^2 = 1\). In polar coordinates, the unit circle can be parameterized as \(x = cos(t)\) and \(y = sin(t)\) with \(t\) ranging from 0 to \(2\pi\). However, since the path is from (1,0) to (0,1), we are interested in the first quadrant where \(t\) ranges from 0 to \(\pi/2\).

Substitute Parameters into the Function

Now you can substitute these parameterized coordinates \(x = cos(t)\) and \(y = sin(t)\) into the function \(f(x,y) = x^2 + y^2\). You get \(f(t) = (cos^2(t) + sin^2(t))\). But since \(cos^2(t) + sin^2(t) = 1\), the function simplifies to \(f(t) = 1\).\nAlso, the differential \(ds\) for a curve given in polar form \(ds = r dt\), for a unit circle \(r=1\), so \(ds = dt\). So the line integral becomes: \(\int_{C} f(t) dt\).

Evaluate the Line Integral

To evaluate the line integral, you need to substitute the function \(f(t)\) into the integral and set the limits from 0 to \(\pi/2\). So the line integral is evaluated as: \(\int_{0}^{\pi/2} 1 dt = t|_0^{\pi/2} = \pi/2 - 0 = \pi/2\).