Math Problem Statement
If C is the part of the circle (x4)2+(y4)2=1 in the first quadrant, find the following line integral with respect to arc length.
∫C(9x−8y)ds
Solution
We are asked to find the line integral of , where is the part of the circle defined by the equation that lies in the first quadrant. Here, represents the differential arc length, and the circle is centered at with a radius of 1.
Step 1: Parametrize the Circle
The equation of the circle can be written as: We can parametrize this circle in terms of an angle , where since we are only interested in the first quadrant: where .
Step 2: Compute the Arc Length Element
The arc length element is given by: First, compute the derivatives of and : Now, substitute into the expression for : So, .
Step 3: Set Up the Line Integral
The line integral becomes: Simplifying the integrand: Thus, the integral becomes:
Step 4: Evaluate the Integral
We can now evaluate the integral term by term:
Step 5: Final Answer
Now, sum the results of the integrals:
Thus, the value of the line integral is:
Would you like more details or explanations about any step?
Here are 5 related questions you might explore:
- How would the result change if the circle were centered at a different point?
- Can the same approach be used for ellipses, and how?
- What is the significance of the line integral in physical contexts?
- How does the parametrization change for different sections of the circle?
- Can this integral be solved in polar coordinates directly?
Tip: When solving line integrals, always ensure that the limits of the parameterization correctly represent the section of the curve being integrated over.
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Math Problem Analysis
Mathematical Concepts
Line Integrals
Parametric Equations
Calculus
Arc Length
Trigonometry
Formulas
Equation of a circle: (x - h)^2 + (y - k)^2 = r^2
Parametric equations: x(θ) = h + cos(θ), y(θ) = k + sin(θ)
Arc length element: ds = sqrt((dx/dθ)^2 + (dy/dθ)^2) dθ
Line integral: ∫C f(x, y) ds
Theorems
Pythagorean Theorem
Fundamental Theorem of Line Integrals
Suitable Grade Level
Undergraduate Calculus (Grades 11-12 and above)
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