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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.
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