Question

In Exercises 13-16, evaluate \(\int_{C}(x+4 \sqrt{y}) d s\) along the given path. \(C:\) line from (0,0) to (1,1)

Short answer

The result of the line integral \(\int_{C}(x+4 \sqrt{y}) d s\) along the path C, which is the line from (0,0) to (1,1), is \(\sqrt{2}+ \frac{8\sqrt{2}}{3}\).

Step-by-step solution

Parameterization of the path C

A parameterization of the line from (0,0) to (1,1) can be done using a parameter \(t\) that ranges from 0 to 1. This gives us the parameterized functions \(x(t) = t\) and \(y(t) = t\).

Calculating the differentials

For this particular integral, we need the differential of the parameterized path. The differentials \(dx\) and \(dy\) can be calculated by taking derivatives of \(x(t)\) and \(y(t)\) with respect to \(t\), and then multiplying by the differential \(dt\). The differential \(ds\) can be obtained by using the formula \(ds= \sqrt {(dx)^2 + (dy)^2} dt\) . For our parameterized path, \(dx/dt=dy/dt= 1\) , so \(ds= \sqrt {1+1} dk = sqrt(2) dt \).

Substituting into the integral

Substitute the parameterizations \(x(t)= t\) and \(y(t)= t\), and \(ds\) into the integral. This changes the line integral into an ordinary integral over \(t\), which can be computed by property of integrals: \(\int_{C}(x+4 \sqrt{y}) d s = \int_{0}^{1}(t + 4\sqrt{t})\sqrt{2} dt \).

Evaluating the integral

Integrating yields \(\sqrt{2}*\frac{1}{2}*t^2+4*\frac{2}{3}*t^{1.5}|_{0}^{1}\). Which simplifies to \(\sqrt{2}*\frac{1}{2}+4*\sqrt{2}*\frac{2}{3} = \sqrt{2}+ \frac{8\sqrt{2}}{3}\).