Question

Evaluate the integral \(\int_{C}(2 x-y) d x+(x+3 y) d y\) along the path \(C\). \(C:\) elliptic path \(x=4 \sin t, y=3 \cos t\) from (0,3) to (4,0)

Short answer

The integral over the path from (0,3) to (4,0) is \(\frac{5\pi}{4}-2\).

Step-by-step solution

Parameterization of Path

The given path is an elliptic path given in parametric form \(x=4 \sin t, y=3 \cos t\). The limits for \(t\) can be found from the points (0,3) to (4,0), which translates to \(t\) moving from \(\frac{\pi}{2}\) to \(0\), because when \(t= \frac{\pi}{2}\), \(x=0, y=3\) and when \(t=0\), \(x=4, y=0\).

Replacing \(x\) and \(y\) Expressions

Replace \(x\) and \(y\) in the integrand with their expressions in terms of \(t\): \((2 x-y) dx + (x+3 y) dy = (2*4\sin t -3\cos t) d(4\sin t) + ((4\sin t) + 3*3\cos t) d(3\cos t)\).

Computing the Differential \(dx\) and \(dy\)

The differentials are computed as \(d(4\sin t) = 4\cos t dt\) and \(d(3\cos t) = -3\sin t dt\). Substituting these, the integral becomes: \(\int_{C}(2*4\sin t -3\cos t)*4\cos t dt + ((4\sin t) + 3*3\cos t)*(-3\sin t) dt\).

Simplifying Integral

Simplify the integral to \(-\int_{\frac{\pi}{2}}^{0} [(32\sin t \cos t -12\cos^2 t -12\sin^2 t - 27\sin t \cos t)\] dt = \int_{0}^{\frac{\pi}{2}} [32\sin t \cos t - 12\cos^2 t -12\sin^2 t + 27\sin t \cos t] dt\).

Evaluating the Integral

The integral can now be evaluated using integration techniques yielding \(32*\frac{1}{4} + 0 - 6*\frac{\pi}{2} - 27*\frac{1}{4}\) when the limits are substituted. Adding these up results to \(\frac{5\pi}{4}-2\).