Using Green's Theorem, calculate the area of the indicated region.
-The area bounded above by $y = 8$ and below by $y = \frac { 8 } { 25 } x ^ { 2 }$
A) $\frac { 320 } { 3 }$
B) 0
C) $\frac { 160 } { 3 }$
D) $\frac { 80 } { 3 }$

Question

Quizplus · Accepted Answer

To apply Green's Theorem, we need to convert the given region into a closed curve. The curve can be obtained by solving the equation for $y$ in terms of $x$, which gives $y = \frac { 8 } { 25 } x ^ { 2 }$. The curve starts at $x = 0$ and ends at $x = 5$, and the orientation is counterclockwise.

The line integral of a vector field $\mathbf { F } = P \mathbf { i } + Q \mathbf { j }$ around this curve is given by

$$\oint _ { C } \mathbf { F } \cdot d \mathbf { r } = \int _ { 0 } ^ { 5 } \left( Q \frac { dy } { dx } - P \right) dx + \int _ { \frac { 8 } { 25 } \cdot 5 ^ { 2 } } ^ { 8 } \left( - \frac { P } { 2 y } \right) dy + \int _ { 5 } ^ { 0 } \left( P \frac { dy } { dx } - Q \right) dx + \int _ { 8 } ^ { \frac { 8 } { 25 } \cdot 0 ^ { 2 } } \left( \frac { P } { 2 y } \right) dy$$

We can simplify this expression by noting that the second and fourth integrals are equal in magnitude but opposite in sign, and cancel each other. Similarly, the first and third integrals are equal in magnitude but opposite in sign, and cancel each other. Therefore, we are left with only one integral:

$$\oint _ { C } \mathbf { F } \cdot d \mathbf { r } = \int _ { \frac { 8 } { 25 } \cdot 5 ^ { 2 } } ^ { 8 } \left( - \frac { P } { 2 y } \right) dy$$

To apply Green's Theorem, we need to find a vector field $\mathbf { F }$ such that $Q \frac { dy } { dx } - P = 1$. This condition is satisfied by $\mathbf { F } = - y \mathbf { i } + \frac { x } { 2 } \mathbf { j }$. Therefore, we have

$$\int _ { \frac { 8 } { 25 } \cdot 5 ^ { 2 } } ^ { 8 } \left( - \frac { P } { 2 y } \right) dy = \iint _ { R } \left( \frac { \partial Q } { \partial x } - \frac { \partial P } { \partial y } \right) dA = \iint _ { R } 1 dA$$

where $R$ is the region enclosed by the curve. Thus, the area of $R$ is given by

$$\iint _ { R } 1 dA = \int _ { 0 } ^ { 5 } \int _ { \frac { 8 } { 25 } x ^ { 2 } } ^ { 8 } dydx = \frac { 320 } { 3 }$$

Therefore, the answer is (C).

Using Green's Theorem, Calculate the Area of the Indicated Region y=8y = 8y=8

Using Green's Theorem, Calculate the Area of the Indicated Region $y = 8$