Let \vec F = 6xe^y \vec i + 3x^2e^y \vec j and \vec G = 6(x - y) \vec i + 3(x + y) \vec j....

Question:

Let {eq}\vec F = 6xe^y \vec i + 3x^2e^y \vec j {/eq} and {eq}\vec G = 6(x - y) \vec i + 3(x + y) \vec j. {/eq} Let be the path consisting of lines from (0, 0) to (3, 0) to (3, 2) to (0, 0). Find each of the following integrals exactly:

{eq}(a) \int_C \vec F \cdot d \vec r = \\ (b) \int_C \vec G \cdot d \vec r = {/eq}

Green's Therorem:

We will use Green's Theorem {eq}\oint F_{1}dx+F_{2}dy=\int \int \left ( \frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y} \right )dxdy {/eq} to solve the problem.

Answer and Explanation:

To solve the problem we will use Green's Theorem:

{eq}\oint F_{1}dx+F_{2}dy=\int \int \left ( \frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y} \right )dxdy {/eq}

Now let us find the partial derivatives:

{eq}\frac{\partial F_{2}}{\partial x}=6xe^{y}\\ \frac{\partial F_{1}}{\partial y}=6xe^{y} {/eq}

The integral will be:

{eq}=\int \int 0dxdy\\ =0 {/eq}

b) To solve the problem we will use Green's Theorem:

{eq}\oint F_{1}dx+F_{2}dy=\int \int \left ( \frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y} \right )dxdy {/eq}

Now let us find the partial derivatives:

{eq}\frac{\partial F_{2}}{\partial x}=3\\ \frac{\partial F_{1}}{\partial y}=-6 {/eq}

The integral will be:

{eq}=\int \int 0dxdy\\ =9dxdy\\ =9\times 3\times 2 {/eq}


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The Fundamental Theorem of Calculus

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