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sesenic [268]
4 years ago
14

Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices

(0, 0), (4, 0), (4, 5), (0, 5) (a) directly (b) using Green's Theorem

Mathematics
1 answer:
Jet001 [13]4 years ago
8 0
Answer to question 1

The given line integral is \int\limits_C {xydx+x^2dy} \,

We evaluate the first line integral from (0,0) to (4,0).

An equation of the straight line joining (0,0) and (4,0) in the xy plane is y=0.

\Rightarrow dy=0

The first line integral now becomes,

l_1=\int\limits^4_0 {x(0)dx+x^2(0)} \,=0

We evaluate the second line integral from (4,0) to (4,5).

An equation of the straight line joining (4,0) and (4,5) in the xy plane is x=4.

\Rightarrow dx=0

The second line integral now becomes,

l_2=\int\limits^5_0 {(4)y(0)+4^2dy} \,=0

l_2=\int\limits^5_0 {16dy} \,=80

We now evaluate the third line integral from (4,5) to (0,5).

An equation of the straight line joining (4,5) and (0,5) in the xy plane is y=5.

\Rightarrow dy=0

The third line integral now becomes,

l_3=\int\limits^0_4 {x(5)dx+x^2(0)} \,

l_3=\int\limits^0_4 {5xdx} \,=-40

We now evaluate the fourth line integral from (0,5) to (0,0).

An equation of the straight line joining (0,5) and (0,0) in the xy plane is x=0.

\Rightarrow dx=0

The fourth line integral now becomes,

l_4=\int\limits^0_5 {(0)y(0)+0^2dy} \,=0

We now add all the line integrals to get,

l=0+80+-40+0=40

Answer to question 2

According to the Green's Theorem,

\int\limits_C {P(x,y)dx+Q(x,y)dy} \,=\int\limits \, \int\limits_D ({\frac{\partial Q}{\partial x}}-{\frac{\partial P}{\partial y}}) \,dA.

This implies that,

\int\limits_C {xydx+x^2dy} \,=\int\limits^5_0 \, \int\limits^4_0 ({2x-x }) \, dxdy

This simplifies to  

\int\limits_C {xydx+x^2dy} \,=\int\limits^5_0 \, \int\limits^4_0 ({x }) \, dxdy

We evaluate the inner integral to get,

\int\limits_C {xydx+x^2dy} \,=\int\limits^5_0 \, ({8 }) \, dy

We now integrate again, to obtain,

\int\limits_C {xydx+x^2dy} \,=8\times 5=40 \,

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