## 1. 1 (8 points). Calculate the area of the quadrilateral with vertices at (0,0), (4,2), (2,2), and (4,6) An explicit line integral or sum of line integrals equal to this area is:_________________________________ The area is: _______________________________

Solution. This can be done with Green's formula:

Area = Integral[(-y) dx} counterwise along the perimeter of the figure. The
given points are listed in counterclockwise order,

On the first leg, we can take x=4 t, y = 2 t, 0 <= t <= 1:

In[1]:=

`  Int1 = Integrate[(-2 t)*4, {t,0,1}]`

Out[1]=

```  -4
```

On the second leg, we can take x=4 - 2 t, y = 2, 0 <= t <= 1:

In[2]:=

`  Int2 = Integrate[(-2)*(-2), {t,0,1}]`

Out[2]=

```  4
```

On the third leg, we can take x=2 t + 2, y = 2+ 4 t, 0 <= t <= 1:

In[3]:=

`  Int3 = Integrate[(-2 - 4 t)*(2), {t,0,1}]`

Out[3]=

```  -8
```

On the fourth leg, we return to the origin and can take x=4 - 4 t, y = 6 - 6 t, 0 <= t <= 1:

In[4]:=

`  Int4 = Integrate[(-6 + 6 t)*(-4), {t,0,1}]`

Out[4]=

```  12
```

In[5]:=

`  Area = Int1 + Int2 + Int3 + Int4`

Out[5]=

```  4
```

The explicit expression asked for is
Area = Integrate[(-2 t)*4, {t,0,1}]
+ Integrate[(-2)*(-2), {t,0,1}]
+ Integrate[(-2 - 4 t)*(2), {t,0,1}]
+Integrate[(-6 + 6 t)*(-4), {t,0,1}],
and its value is 4. You could equally well calculate the integral of x dy.

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