Green's Theorem - AP Calculus BC

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Question

Use Green's Theorem to evaluate $\oint_C 5xy \ dx +x^3y^2 \ dy$ , where is a triangle with vertices , , with positive orientation.

Answer

First we need to make sure that the conditions for Green's Theorem are met.

The conditions are met because it is positively oriented, piecewise smooth, simple, and closed under the region (see below).

$0\leq x \leq 1$

$0 \leq y \leq 2x$

In this particular case , and , where , and refer to $\oint_C P \ dx +Q \ dy$ .

We know from Green's Theorem that

$\oint_C P \ dx +Q \ dy=\int \int_D \Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big) dA$

So lets find the partial derivatives.

$\frac{\partial Q}{\partial x}=3x^2y^2$

$\frac{\partial P}{\partial y}=5x$

$\oint_C 5xy \ dx +x^3y^2 \ dy=\int \int_D 3x^2y^2-5x \ dA$

$=\int_{0}^{1} \int_{0}^{2x} 3x^2y^2-5x \ dy\ dx$

$=\int_{0}^{1} \Big(\frac{1}{3}\cdot3x^2y^3-5xy\Big)\Big|_{0}^{2x}\ dx$

$=\int_{0}^{1} \Big(x^2(2x)^3-5x(2x)\Big)\ dx$

$=\int_{0}^{1} 8x^5-10x^2\ dx$

$= \frac{4}{3}x^6-\frac{10}{3}x^3\Big|_{0}^{1}$

$= \frac{4}{3}(1)^6-\frac{10}{3}(1)^3-0=\frac{4}{3}-\frac{10}{3}=-\frac{6}{3}=-2$

Compare your answer with the correct one above

Question

Use Green's Theorem to evaluate the line integral

$\int_{c}3ydx-4xdy$

over the region R, described by connecting the points , orientated clockwise.

Answer

Using Green's theorem

$\int_C{F}dr=\int \int\left[{\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}}\right]dA$

$\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}=-4-3=-7$

$Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

since the region is oriented clockwise, we would have

$\int_C{F}dr=-\int\int\left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

which gives us

$\int_C{F} dr=-(-7)\times4=28$

Compare your answer with the correct one above

Question

Use Greens Theorem to evaluate the line integral

$\int_C 5y dx - 4x dy$

over the region connecting the points oriented clockwise

Answer

$\int_C 5y dx - 4x dy$

Using Green's theorem

$\int_C F dr=\int\int \left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

$Q=-4x\ \frac{\partial{Q}}{\partial{x}}=-4$ $P=5y\ \frac{\partial{P}}{\partial{y}}=5$

$Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

Since the region is oriented clockwise

$\int_C F dr=-\int\int (-4-5)dA$

$\int_C F dr=\int\int 9dA=(9)(4)=36$

Compare your answer with the correct one above

Question

Use Green's Theorem to evaluate $\oint_C 5xy \ dx +x^3y^2 \ dy$ , where is a triangle with vertices , , with positive orientation.

Answer

First we need to make sure that the conditions for Green's Theorem are met.

The conditions are met because it is positively oriented, piecewise smooth, simple, and closed under the region (see below).

$0\leq x \leq 1$

$0 \leq y \leq 2x$

In this particular case , and , where , and refer to $\oint_C P \ dx +Q \ dy$ .

We know from Green's Theorem that

$\oint_C P \ dx +Q \ dy=\int \int_D \Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big) dA$

So lets find the partial derivatives.

$\frac{\partial Q}{\partial x}=3x^2y^2$

$\frac{\partial P}{\partial y}=5x$

$\oint_C 5xy \ dx +x^3y^2 \ dy=\int \int_D 3x^2y^2-5x \ dA$

$=\int_{0}^{1} \int_{0}^{2x} 3x^2y^2-5x \ dy\ dx$

$=\int_{0}^{1} \Big(\frac{1}{3}\cdot3x^2y^3-5xy\Big)\Big|_{0}^{2x}\ dx$

$=\int_{0}^{1} \Big(x^2(2x)^3-5x(2x)\Big)\ dx$

$=\int_{0}^{1} 8x^5-10x^2\ dx$

$= \frac{4}{3}x^6-\frac{10}{3}x^3\Big|_{0}^{1}$

$= \frac{4}{3}(1)^6-\frac{10}{3}(1)^3-0=\frac{4}{3}-\frac{10}{3}=-\frac{6}{3}=-2$

Compare your answer with the correct one above

Question

Use Green's Theorem to evaluate the line integral

$\int_{c}3ydx-4xdy$

over the region R, described by connecting the points , orientated clockwise.

Answer

Using Green's theorem

$\int_C{F}dr=\int \int\left[{\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}}\right]dA$

$\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}=-4-3=-7$

$Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

since the region is oriented clockwise, we would have

$\int_C{F}dr=-\int\int\left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

which gives us

$\int_C{F} dr=-(-7)\times4=28$

Compare your answer with the correct one above

Question

Use Greens Theorem to evaluate the line integral

$\int_C 5y dx - 4x dy$

over the region connecting the points oriented clockwise

Answer

$\int_C 5y dx - 4x dy$

Using Green's theorem

$\int_C F dr=\int\int \left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

$Q=-4x\ \frac{\partial{Q}}{\partial{x}}=-4$ $P=5y\ \frac{\partial{P}}{\partial{y}}=5$

$Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

Since the region is oriented clockwise

$\int_C F dr=-\int\int (-4-5)dA$

$\int_C F dr=\int\int 9dA=(9)(4)=36$

Compare your answer with the correct one above

Question

Use Green's Theorem to evaluate $\oint_C 5xy \ dx +x^3y^2 \ dy$ , where is a triangle with vertices , , with positive orientation.

Answer

First we need to make sure that the conditions for Green's Theorem are met.

The conditions are met because it is positively oriented, piecewise smooth, simple, and closed under the region (see below).

$0\leq x \leq 1$

$0 \leq y \leq 2x$

In this particular case , and , where , and refer to $\oint_C P \ dx +Q \ dy$ .

We know from Green's Theorem that

$\oint_C P \ dx +Q \ dy=\int \int_D \Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big) dA$

So lets find the partial derivatives.

$\frac{\partial Q}{\partial x}=3x^2y^2$

$\frac{\partial P}{\partial y}=5x$

$\oint_C 5xy \ dx +x^3y^2 \ dy=\int \int_D 3x^2y^2-5x \ dA$

$=\int_{0}^{1} \int_{0}^{2x} 3x^2y^2-5x \ dy\ dx$

$=\int_{0}^{1} \Big(\frac{1}{3}\cdot3x^2y^3-5xy\Big)\Big|_{0}^{2x}\ dx$

$=\int_{0}^{1} \Big(x^2(2x)^3-5x(2x)\Big)\ dx$

$=\int_{0}^{1} 8x^5-10x^2\ dx$

$= \frac{4}{3}x^6-\frac{10}{3}x^3\Big|_{0}^{1}$

$= \frac{4}{3}(1)^6-\frac{10}{3}(1)^3-0=\frac{4}{3}-\frac{10}{3}=-\frac{6}{3}=-2$

Compare your answer with the correct one above

Question

Use Green's Theorem to evaluate the line integral

$\int_{c}3ydx-4xdy$

over the region R, described by connecting the points , orientated clockwise.

Answer

Using Green's theorem

$\int_C{F}dr=\int \int\left[{\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}}\right]dA$

$\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}=-4-3=-7$

$Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

since the region is oriented clockwise, we would have

$\int_C{F}dr=-\int\int\left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

which gives us

$\int_C{F} dr=-(-7)\times4=28$

Compare your answer with the correct one above

Question

Use Greens Theorem to evaluate the line integral

$\int_C 5y dx - 4x dy$

over the region connecting the points oriented clockwise

Answer

$\int_C 5y dx - 4x dy$

Using Green's theorem

$\int_C F dr=\int\int \left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

$Q=-4x\ \frac{\partial{Q}}{\partial{x}}=-4$ $P=5y\ \frac{\partial{P}}{\partial{y}}=5$

$Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

Since the region is oriented clockwise

$\int_C F dr=-\int\int (-4-5)dA$

$\int_C F dr=\int\int 9dA=(9)(4)=36$

Compare your answer with the correct one above

Question

Use Green's Theorem to evaluate $\oint_C 5xy \ dx +x^3y^2 \ dy$ , where is a triangle with vertices , , with positive orientation.

Answer

First we need to make sure that the conditions for Green's Theorem are met.

The conditions are met because it is positively oriented, piecewise smooth, simple, and closed under the region (see below).

$0\leq x \leq 1$

$0 \leq y \leq 2x$

In this particular case , and , where , and refer to $\oint_C P \ dx +Q \ dy$ .

We know from Green's Theorem that

$\oint_C P \ dx +Q \ dy=\int \int_D \Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big) dA$

So lets find the partial derivatives.

$\frac{\partial Q}{\partial x}=3x^2y^2$

$\frac{\partial P}{\partial y}=5x$

$\oint_C 5xy \ dx +x^3y^2 \ dy=\int \int_D 3x^2y^2-5x \ dA$

$=\int_{0}^{1} \int_{0}^{2x} 3x^2y^2-5x \ dy\ dx$

$=\int_{0}^{1} \Big(\frac{1}{3}\cdot3x^2y^3-5xy\Big)\Big|_{0}^{2x}\ dx$

$=\int_{0}^{1} \Big(x^2(2x)^3-5x(2x)\Big)\ dx$

$=\int_{0}^{1} 8x^5-10x^2\ dx$

$= \frac{4}{3}x^6-\frac{10}{3}x^3\Big|_{0}^{1}$

$= \frac{4}{3}(1)^6-\frac{10}{3}(1)^3-0=\frac{4}{3}-\frac{10}{3}=-\frac{6}{3}=-2$

Compare your answer with the correct one above

Question

Use Green's Theorem to evaluate the line integral

$\int_{c}3ydx-4xdy$

over the region R, described by connecting the points , orientated clockwise.

Answer

Using Green's theorem

$\int_C{F}dr=\int \int\left[{\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}}\right]dA$

$\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}=-4-3=-7$

$Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

since the region is oriented clockwise, we would have

$\int_C{F}dr=-\int\int\left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

which gives us

$\int_C{F} dr=-(-7)\times4=28$

Compare your answer with the correct one above

Question

Use Greens Theorem to evaluate the line integral

$\int_C 5y dx - 4x dy$

over the region connecting the points oriented clockwise

Answer

$\int_C 5y dx - 4x dy$

Using Green's theorem

$\int_C F dr=\int\int \left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

$Q=-4x\ \frac{\partial{Q}}{\partial{x}}=-4$ $P=5y\ \frac{\partial{P}}{\partial{y}}=5$

$Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

Since the region is oriented clockwise

$\int_C F dr=-\int\int (-4-5)dA$

$\int_C F dr=\int\int 9dA=(9)(4)=36$

Compare your answer with the correct one above

Question

Use Green's Theorem to evaluate $\oint_C 5xy \ dx +x^3y^2 \ dy$ , where is a triangle with vertices , , with positive orientation.

Answer

First we need to make sure that the conditions for Green's Theorem are met.

The conditions are met because it is positively oriented, piecewise smooth, simple, and closed under the region (see below).

$0\leq x \leq 1$

$0 \leq y \leq 2x$

In this particular case , and , where , and refer to $\oint_C P \ dx +Q \ dy$ .

We know from Green's Theorem that

$\oint_C P \ dx +Q \ dy=\int \int_D \Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big) dA$

So lets find the partial derivatives.

$\frac{\partial Q}{\partial x}=3x^2y^2$

$\frac{\partial P}{\partial y}=5x$

$\oint_C 5xy \ dx +x^3y^2 \ dy=\int \int_D 3x^2y^2-5x \ dA$

$=\int_{0}^{1} \int_{0}^{2x} 3x^2y^2-5x \ dy\ dx$

$=\int_{0}^{1} \Big(\frac{1}{3}\cdot3x^2y^3-5xy\Big)\Big|_{0}^{2x}\ dx$

$=\int_{0}^{1} \Big(x^2(2x)^3-5x(2x)\Big)\ dx$

$=\int_{0}^{1} 8x^5-10x^2\ dx$

$= \frac{4}{3}x^6-\frac{10}{3}x^3\Big|_{0}^{1}$

$= \frac{4}{3}(1)^6-\frac{10}{3}(1)^3-0=\frac{4}{3}-\frac{10}{3}=-\frac{6}{3}=-2$

Compare your answer with the correct one above

Question

Use Green's Theorem to evaluate the line integral

$\int_{c}3ydx-4xdy$

over the region R, described by connecting the points , orientated clockwise.

Answer

Using Green's theorem

$\int_C{F}dr=\int \int\left[{\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}}\right]dA$

$\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}=-4-3=-7$

$Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

since the region is oriented clockwise, we would have

$\int_C{F}dr=-\int\int\left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

which gives us

$\int_C{F} dr=-(-7)\times4=28$

Compare your answer with the correct one above

Question

Use Greens Theorem to evaluate the line integral

$\int_C 5y dx - 4x dy$

over the region connecting the points oriented clockwise

Answer

$\int_C 5y dx - 4x dy$

Using Green's theorem

$\int_C F dr=\int\int \left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

$Q=-4x\ \frac{\partial{Q}}{\partial{x}}=-4$ $P=5y\ \frac{\partial{P}}{\partial{y}}=5$

$Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

Since the region is oriented clockwise

$\int_C F dr=-\int\int (-4-5)dA$

$\int_C F dr=\int\int 9dA=(9)(4)=36$

Compare your answer with the correct one above

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