Question

${\mathbf{\int }}_{\mathbf{C}}\mathbf{(}\mathit{y}{\mathit{e}}^{\mathbf{x}}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{d}\mathit{x}\mathbf{+}{\mathit{e}}^{\mathbf{x}}\mathit{d}\mathit{y}$,where Cis the semicircle through$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{10}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{10}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{10}\mathbf{)}$(Compare Problem 4.)

Short answer

The solution to this problem is 20

Step-by-step solution

Given Information.

The given information is ${\int }_C(y{e}^x-1)dx+e^{x}dy$.

Definition of Green’s Theorem.

The Green's theorem connects a line integral around a simple closed curve C to a double integral over the plane region D circumscribed by C in vector calculus. Stokes' theorem has a two-dimensional special case.

Find the solution.

Consider an equation and name it as equation 1

${\int }_C(y{e}^x-1)dx+e^{x}dy.......\left(1\right)$

Compare equation (1) to Green’s Theorem.

$$\begin{gathered} P=y{e}^x-1 \\ Q=e^x \end{gathered}$$

Find the differentiation with respect to.

$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0$

This implies that the surface integral will also be zero, which further leads to the line integral along C is negative that of line integral moving from

$(0,10)to(0,-10).$

Let the value be as follows:

$$\begin{gathered} x=0 \\ dx=0 \end{gathered}$$

Find the Integral.

$$\begin{gathered} l={\int }_{10}^{-10}dy \\ =-\left(-10-10\right) \\ =20 \end{gathered}$$

Hence, the solution to this problem is 20.