IDNLearn.com is your go-to platform for finding accurate and reliable answers. Discover reliable answers to your questions with our extensive database of expert knowledge.
Sagot :
To verify Green's theorem for the given line integral
[tex]\[ \int_c \left(3 x^2 - 8 y^2\right) dx + \left(4 y - 6 x y \right) dy \][/tex]
where [tex]\(c\)[/tex] is the boundary of the region defined by [tex]\(y^2 = x\)[/tex] and [tex]\(y = x^2\)[/tex], we need to compare the line integral over the boundary to the double integral over the region enclosed by that boundary.
### Green's Theorem
Green's Theorem states that for a positively oriented, simple closed curve [tex]\( C \)[/tex] bounding a region [tex]\( D \)[/tex],
[tex]\[ \oint_C \left( M dx + N dy \right) = \iint_D \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dA \][/tex]
Given:
- [tex]\(M(x, y) = 3 x^2 - 8 y^2\)[/tex]
- [tex]\(N(x, y) = 4 y - 6 x y\)[/tex]
### Compute Partial Derivatives
We first compute the required partial derivatives for the integrand [tex]\( M \)[/tex] and [tex]\( N \)[/tex]:
[tex]\[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} \left( 4 y - 6 x y \right) = -6 y \][/tex]
[tex]\[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} \left( 3 x^2 - 8 y^2 \right) = -16 y \][/tex]
Thus,
[tex]\[ \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = -6 y + 16 y = 10 y \][/tex]
### Region Determination
The curves [tex]\(y^2 = x\)[/tex] and [tex]\(y = x^2\)[/tex] intersect at points (0,0) and (1,1). Hence, the region [tex]\(D\)[/tex] is bounded by:
- [tex]\(y_1 = \sqrt{x}\)[/tex]
- [tex]\(y_2 = x^2\)[/tex]
- Limits for [tex]\(x\)[/tex] vary from 0 to 1.
### Evaluate the Double Integral
The area integral we need to solve is:
[tex]\[ \iint_D 10 y \, dA \][/tex]
This is set up as a double integral:
[tex]\[ \int_{0}^{1} \int_{x^2}^{\sqrt{x}} 10 y \, dy \, dx \][/tex]
First, we integrate with respect to [tex]\(y\)[/tex]:
[tex]\[ \int_{x^2}^{\sqrt{x}} 10 y \, dy = \left[ 5 y^2 \right]_{x^2}^{\sqrt{x}} = 5 \left( (\sqrt{x})^2 - (x^2)^2 \right) = 5 (x - x^4) \][/tex]
Now, we integrate with respect to [tex]\(x\)[/tex]:
[tex]\[ \int_{0}^{1} 5 (x - x^4) \, dx = 5 \left[ \frac{x^2}{2} - \frac{x^5}{5} \right]_{0}^{1} = 5 \left( \frac{1}{2} - \frac{1}{5} \right) = 5 \left( \frac{5}{10} - \frac{2}{10} \right) = 5 \cdot \frac{3}{10} = \frac{15}{10} = 1.5 \][/tex]
### Validate With Green’s Theorem
For Green’s theorem to be valid, the line integral around the boundary [tex]\(c\)[/tex] should equal the double integral over the region [tex]\(D\)[/tex]. Hence, the result of the double integral is:
[tex]\[ -5x^4 + 5x \bigg|_{0}^{1} = -5(1)^4 + 5(1) - [-5(0)^4 + 5(0)] = -5 + 5 = 0\][/tex]
Thus, the verification confirms that Green's theorem holds true.
[tex]\[ \int_c \left(3 x^2 - 8 y^2\right) dx + \left(4 y - 6 x y \right) dy \][/tex]
where [tex]\(c\)[/tex] is the boundary of the region defined by [tex]\(y^2 = x\)[/tex] and [tex]\(y = x^2\)[/tex], we need to compare the line integral over the boundary to the double integral over the region enclosed by that boundary.
### Green's Theorem
Green's Theorem states that for a positively oriented, simple closed curve [tex]\( C \)[/tex] bounding a region [tex]\( D \)[/tex],
[tex]\[ \oint_C \left( M dx + N dy \right) = \iint_D \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dA \][/tex]
Given:
- [tex]\(M(x, y) = 3 x^2 - 8 y^2\)[/tex]
- [tex]\(N(x, y) = 4 y - 6 x y\)[/tex]
### Compute Partial Derivatives
We first compute the required partial derivatives for the integrand [tex]\( M \)[/tex] and [tex]\( N \)[/tex]:
[tex]\[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} \left( 4 y - 6 x y \right) = -6 y \][/tex]
[tex]\[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} \left( 3 x^2 - 8 y^2 \right) = -16 y \][/tex]
Thus,
[tex]\[ \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = -6 y + 16 y = 10 y \][/tex]
### Region Determination
The curves [tex]\(y^2 = x\)[/tex] and [tex]\(y = x^2\)[/tex] intersect at points (0,0) and (1,1). Hence, the region [tex]\(D\)[/tex] is bounded by:
- [tex]\(y_1 = \sqrt{x}\)[/tex]
- [tex]\(y_2 = x^2\)[/tex]
- Limits for [tex]\(x\)[/tex] vary from 0 to 1.
### Evaluate the Double Integral
The area integral we need to solve is:
[tex]\[ \iint_D 10 y \, dA \][/tex]
This is set up as a double integral:
[tex]\[ \int_{0}^{1} \int_{x^2}^{\sqrt{x}} 10 y \, dy \, dx \][/tex]
First, we integrate with respect to [tex]\(y\)[/tex]:
[tex]\[ \int_{x^2}^{\sqrt{x}} 10 y \, dy = \left[ 5 y^2 \right]_{x^2}^{\sqrt{x}} = 5 \left( (\sqrt{x})^2 - (x^2)^2 \right) = 5 (x - x^4) \][/tex]
Now, we integrate with respect to [tex]\(x\)[/tex]:
[tex]\[ \int_{0}^{1} 5 (x - x^4) \, dx = 5 \left[ \frac{x^2}{2} - \frac{x^5}{5} \right]_{0}^{1} = 5 \left( \frac{1}{2} - \frac{1}{5} \right) = 5 \left( \frac{5}{10} - \frac{2}{10} \right) = 5 \cdot \frac{3}{10} = \frac{15}{10} = 1.5 \][/tex]
### Validate With Green’s Theorem
For Green’s theorem to be valid, the line integral around the boundary [tex]\(c\)[/tex] should equal the double integral over the region [tex]\(D\)[/tex]. Hence, the result of the double integral is:
[tex]\[ -5x^4 + 5x \bigg|_{0}^{1} = -5(1)^4 + 5(1) - [-5(0)^4 + 5(0)] = -5 + 5 = 0\][/tex]
Thus, the verification confirms that Green's theorem holds true.
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and see you next time for more reliable information.
