Evaluate the line integral of the vector field along the curve , where is the line segment from to . Which of the following is the value of the integral ?
Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
- 12. Techniques of Integration7h 39m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
0. Functions
Introduction to Functions
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Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
Find the area of the region enclosed by one loop of the curve .
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Step 1: Recognize that the problem involves finding the area enclosed by one loop of a polar curve. The formula for the area enclosed by a polar curve is A = (1/2) ∫ r² dθ, where r is the given polar function and the integral is evaluated over the range of θ that corresponds to one loop of the curve.
Step 2: Analyze the curve r = sin(4θ). To determine the range of θ for one loop, observe that the curve completes one loop when r returns to zero after starting at zero. Solve sin(4θ) = 0 to find the values of θ that define the interval for one loop. These values are θ = 0 and θ = π/4.
Step 3: Substitute r = sin(4θ) into the area formula. The integral becomes A = (1/2) ∫[0 to π/4] (sin(4θ))² dθ. This represents the area enclosed by one loop of the curve.
Step 4: Simplify the integrand (sin(4θ))² using the trigonometric identity sin²(x) = (1 - cos(2x))/2. Replace (sin(4θ))² with (1 - cos(8θ))/2 in the integral. The area formula now becomes A = (1/2) ∫[0 to π/4] [(1 - cos(8θ))/2] dθ.
Step 5: Split the integral into two parts: A = (1/4) ∫[0 to π/4] 1 dθ - (1/4) ∫[0 to π/4] cos(8θ) dθ. Evaluate each integral separately. The first integral ∫[0 to π/4] 1 dθ gives the length of the interval, and the second integral ∫[0 to π/4] cos(8θ) dθ can be solved using standard integration techniques for cosine functions.
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