Question

Evaluate the line integral by two methods: (a) directly and (b) using Green's Theorem.

\(\oint_C {(x - y)} dx + (x + y)dy\),

\(C\)is the circle with center the origin and radius\(2\).

Short answer

a) The value of line integral\(\oint_C {(x - y)} dx + (x + y)dy\)is\(\underline {8\pi } \).

b) The value of line integral \(\oint_C {(x - y)} dx + (x + y)dy\)by the use of Green's Theorem is \(8\pi \).

Step-by-step solution

Given data

The line integral is\(\oint_C {(x - y)} dx + (x + y)dy\).

The curve \(C\) is circle with centre origin and radius 2.

The equation of circle and Green’s Theorem

The equation of circle with centre origin is given by,

\({x^2} + {y^2} = {r^2}\)

Green's Theorem:

"Let\(C\)be a positively oriented, piecewise-smooth, simple closed curve in the plane and let\(D\)be the region bounded by\(C\). If\(P\)and\(Q\)have continuous partial derivatives on an open region that contains\(D\), then ."

Find the value of the line integral by direct method

a)

The equation of circle with centre origin is given by,

\({x^2} + {y^2} = {r^2}\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)

Consider parametric equations of curve\(C,0 \le t \le 2\pi \)as,

\(\begin{array}{l}x = 2\cos t\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\\y = 2\sin t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\end{array}\)

Substitute\(2\cos t\)for\(x,2\sin t\)for\(y\), and 2 for\(r\)in equation (1).

\(\begin{align} & {{(2\cos t)}^{2}}+{{(2\sin t)}^{2}} ={{2}^{2}} \\ & 4{{\cos }^{2}}t+4{{\sin }^{2}}t=4 \\ & 4\left( {{\cos }^{2}}t+{{\sin }^{2}}t \right)=4 \\ & 4(1)=4\quad \left\{ \because {{\cos }^{2}}x+{{\sin }^{2}}=1 \right\} \end{align}\)

\(4 = 4\)

The LHS equal to the RHS. Hence, the parametric equations represents the circle\(C\)with origin as centre and radius 2 .

Differentiate equation\((2)\)with respect to\(t\).

\(\begin{align} & \frac{dx}{dt}=\frac{d}{dt}(2\cos t) \\ & \frac{dx}{dt}=2(-\sin t)dx \\ & =-2\sin tdt\quad \left\{ \because \frac{d}{dt}(\cos t)=-\sin t \right\} \end{align}\)

Differentiate equation\((3)\)with respect to\(t\).

\(\begin{align} & \frac{dy}{dt}=\frac{d}{dt}(2\sin t) \\ & \frac{dy}{dt}=2(\cos t) \\ & dy=2\cos tdt\quad \left\{ \because \frac{d}{dt}(\sin t)=\cos t \right\} \\ \end{align}\)

Find the value of line integral\(\oint_C x ydx + {x^2}dy\).

\(\begin{aligned}\oint_C x ydx + {x^2}dy & = \int_0^{2\pi } {(2\cos t - 2\sin t)} ( - 2\sin tdt) + (2\cos t + 2\sin t)(2\cos tdt)\\ & = \int_0^{2\pi } {\left( { - 4\cos t\sin t + 4{{\sin }^2}t + 4{{\cos }^2}t + 4\sin t\cos t} \right)} dt\\ & = \int_0^{2\pi } 4 \left( {{{\sin }^2}t + {{\cos }^2}t} \right)dt\\ & = \int_0^{2\pi } 4 (1)dt\end{aligned}\)

Integrate the equation.

\(\begin{align} & \oint_{C}{(x-y)}dx+(x+y)dy=4[t]_{0}^{2\pi }\quad \left\{ \because \int{d}t=t \right\} \\ & =4(2\pi -0) \\ & =8\pi \end{align}\)

Thus, the value of line integral \(\oint_C {(x - y)} dx + (x + y)dy\) is \(\underline {8\pi } \).

Find the value of the line integral by Green's Theorem

b)

The circle\(C\)is positively oriented, piecewise-smooth, and simply closed curve with domain\(0 \le t \le 2\pi \)and hence Green's theorem is applicable.

Compare the two expressions\(\int_C P dx + Qdy\)and\(\oint_C {(x - y)} dx + (x + y)dy\)and obtain,

\(\begin{aligned}{l}P & = x - y\\Q & = x + y\end{aligned}\)

Find the value of\(\frac{{\partial P}}{{\partial y}}\).

\(\begin{align} \frac{\partial P}{\partial y} & =\frac{\partial }{\partial y}(x-y) \\ & =-1\quad \left\{ \because \frac{\partial }{\partial t}(y)=1 \right\} \end{align}\)

Find the value of \(\frac{{\partial Q}}{{\partial x}}\).

\(\begin{align} \frac{\partial Q}{\partial x} & =\frac{\partial }{\partial x}(x+y) \\ & =1\left\{ \because \frac{\partial }{\partial t}(t)=1 \right\} \end{align}\)

Substitute\(x - y\)for\(P,x + y\)for\(Q, - 1\)for\(\frac{{\partial P}}{{\partial y}}\), and 1 for\(\frac{{\partial Q}}{{\partial x}}\)in the above mentioned formula.

\(\begin{align} & \oint_{C}{(x-y)}dx+(x+y)dy=\iint_{D}{(1+1)}dA \\ & =\iint_{D}{2}dA\oint_{C}{(x-y)}dx+(x+y)dy \\ & =2\left( \iint_{D}{d}A \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 4 \right) \end{align}\)

The area of circle is\(\pi {r^2}\). Hence,

x.\(\iint_{D}{d}A =\pi {{r}^{2}}\)

Substitute 2 for\(r\),

\(\begin{align} \iint_{D}{d}A & =\pi \left( {{2}^{2}} \right) \\ & =4\pi \end{align}\)

Substitute \(4\pi \) for in equation (4).

\(\begin{aligned}\oint_C {(x - y)} dx + (x + y)dy & = 2(4\pi )\\ & = 8\pi \end{aligned}\)

Thus, the value of line integral \(\oint_C {(x - y)} dx + (x + y)dy\) by Green's Theorem is \(\underline {8\pi } \).