Problem 7
Question
Fill in the blank or answer true/false. Where appropriate, assume continuity of \(P, O\), and their first partial derivatives. The integral \(\int_{C}\left(x^{2}+y^{2}\right) d x+2 x y d y\), where \(C\) is given by \(y=x^{3}\) from \((0,0)\) to \((1,1)\), has the same value on the curve \(y=x^{6}\) from \((0,0)\) to \((1,1)\).____
Step-by-Step Solution
Verified Answer
True, the integral has the same value on both curves y=x^3 and y=x^6.
1Step 1: Analyze the Given Line Integral
The integral given is \( \int_{C} (x^{2} + y^{2}) \, dx + 2xy \, dy \), where \( C \) follows the path \( y = x^3 \) from point \( (0,0) \) to \( (1,1) \). This integral can be expressed as \( \int_{C} P \, dx + Q \, dy \) with \( P = x^{2} + y^{2} \) and \( Q = 2xy \).
2Step 2: Apply Green's Theorem
To check if Green's Theorem can be applied, the region bounded by the two paths \( y = x^3 \) and \( y = x^6 \) must satisfy continuity conditions for \( P \) and \( Q \). Green's Theorem states: \[ \oint_{C} P \, dx + Q \, dy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \].
3Step 3: Compute Partial Derivatives
Calculate the partial derivatives: \( \frac{\partial Q}{\partial x} = 2y \) and \( \frac{\partial P}{\partial y} = 2y \). Thus, \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2y - 2y = 0 \).
4Step 4: Evaluate the Area Integral
Substituting \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 \) into the area integral \( \iint_{R} 0 \, dA = 0 \) implies that, according to Green's Theorem, the line integral over any path bounding \( R \) is zero.
5Step 5: Conclude and Fill the Blank
Since the region integral is zero, the line integrals over the paths \( y=x^3 \) and \( y=x^6 \) from \( (0,0) \) to \( (1,1) \) must be equal. Therefore, both integrals have the same value, confirming that the given paths can take any value as long as they start and end at the same points and the integrand conditions are satisfied.
Key Concepts
Line IntegralPartial DerivativesPath Independence
Line Integral
Among various mathematical operations, a line integral allows us to integrate a function along a curve or a path. It is commonly used in vector calculus to find quantities like work done by a force field on a particle moving along a path or the circulation of a fluid along a path. In simple terms, it is an integral that evaluates the sum of values of a function over a path.
For the given integral in the exercise: \( \int_{C} (x^{2} + y^{2}) \, dx + 2xy \, dy \), where \( C \) is defined by certain paths (like \( y = x^3 \) and \( y = x^6 \)), we are integrating the function \( f(x, y) = (x^2 + y^2, 2xy) \) along these paths.
In contexts involving Green's Theorem, such line integrals are often redefined in terms of a vector field \( F = (P, Q) \), with the path being analyzed to see how the integral behaves over different curves between the same start and endpoint.
For the given integral in the exercise: \( \int_{C} (x^{2} + y^{2}) \, dx + 2xy \, dy \), where \( C \) is defined by certain paths (like \( y = x^3 \) and \( y = x^6 \)), we are integrating the function \( f(x, y) = (x^2 + y^2, 2xy) \) along these paths.
In contexts involving Green's Theorem, such line integrals are often redefined in terms of a vector field \( F = (P, Q) \), with the path being analyzed to see how the integral behaves over different curves between the same start and endpoint.
Partial Derivatives
Partial derivatives are used to differentiate multivariable functions with respect to one variable while keeping the others constant. They are crucial in analyzing changes in functions with more than one variable, such as vector fields or surfaces.
In Green’s Theorem, they play a vital role, as seen in the exercise. For example, the function \( Q = 2xy \) has the partial derivative with respect to \( x \), calculated as \( \frac{\partial Q}{\partial x} = 2y \). Similarly, with \( P = x^2 + y^2 \), we compute its partial derivative with respect to \( y \) as \( \frac{\partial P}{\partial y} = 2y \).
These partial derivatives are used to evaluate the expression \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \). In this case, the expression yields \( 0 \), which simplifies the application of Green's Theorem by indicating no net "curl" or circulation within the region enclosed by the paths.
In Green’s Theorem, they play a vital role, as seen in the exercise. For example, the function \( Q = 2xy \) has the partial derivative with respect to \( x \), calculated as \( \frac{\partial Q}{\partial x} = 2y \). Similarly, with \( P = x^2 + y^2 \), we compute its partial derivative with respect to \( y \) as \( \frac{\partial P}{\partial y} = 2y \).
These partial derivatives are used to evaluate the expression \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \). In this case, the expression yields \( 0 \), which simplifies the application of Green's Theorem by indicating no net "curl" or circulation within the region enclosed by the paths.
Path Independence
Path independence in the context of line integrals and vector fields implies that the value of the integral depends only on the starting and endpoint, not the specific path taken between them.
This concept is tied closely to the notion of a conservative vector field, where the line integral of a field over any closed path is zero.
From the exercise, using Green’s Theorem, we determined that the line integral over the enclosed area, i.e., the difference between the paths \( y = x^3 \) and \( y = x^6 \), is zero because \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 \).
This finding confirms that the integral is path-independent between points \((0,0)\) and \((1,1)\) as long as the conditions of continuity are met, resulting in the same integral value across different paths with the same endpoints.
This concept is tied closely to the notion of a conservative vector field, where the line integral of a field over any closed path is zero.
From the exercise, using Green’s Theorem, we determined that the line integral over the enclosed area, i.e., the difference between the paths \( y = x^3 \) and \( y = x^6 \), is zero because \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 \).
This finding confirms that the integral is path-independent between points \((0,0)\) and \((1,1)\) as long as the conditions of continuity are met, resulting in the same integral value across different paths with the same endpoints.
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