100- x2 dx 8.4.9 Let r = 10 sin 0 so that d.r = 10 cos 0 de. Note that v/100 – x2 = 10 cos 0. Then T/3 V3 V3 257 T/3 cos? 0 de = 50 T/3 (1+ cos 20) d0 = 50 ( 0+ 7/6 sin 20 100 = 50 2 4 4
100- x2 dx 8.4.9 Let r = 10 sin 0 so that d.r = 10 cos 0 de. Note that v/100 – x2 = 10 cos 0. Then T/3 V3 V3 257 T/3 cos? 0 de = 50 T/3 (1+ cos 20) d0 = 50 ( 0+ 7/6 sin 20 100 = 50 2 4 4
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Hi, my question is how did we get the new limits pi/3 & pi/6 (circled in pink) . What did we plug the old limits into? And how was it derived. Thank you.
![The text is a mathematical solution to a definite integral problem, involving trigonometric substitution. It is appropriate for an educational website explaining calculus concepts.
**Problem Description:**
- **Given Integral:** \(\int_{5}^{5\sqrt{3}} \sqrt{100 - x^2} \, dx\)
**Solution Steps:**
1. **Substitution:**
- Let \(x = 10 \sin \theta\), which implies \(dx = 10 \cos \theta \, d\theta\).
- Notice that \(\sqrt{100 - x^2} = 10 \cos \theta\).
2. **Changing Limits of Integration:**
- When \(x = 5\), \(\theta = \frac{\pi}{6}\).
- When \(x = 5\sqrt{3}\), \(\theta = \frac{\pi}{3}\).
3. **Integral Transformation:**
- The integral becomes \(\int_{\pi/6}^{\pi/3} 100 \cos^2 \theta \, d\theta\).
4. **Use of Trigonometric Identity:**
- Convert \(\cos^2 \theta\) using the identity \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\).
5. **Integral Evaluation:**
- Compute \(\int_{\pi/6}^{\pi/3} 50 (1 + \cos 2\theta) \, d\theta\).
- The result simplifies to \(50 \left[ \theta + \frac{\sin 2\theta}{2} \right]_{\pi/6}^{\pi/3}\).
6. **Final Calculation:**
- Perform the final computation to get \(50 \left(\frac{\pi}{3} + \frac{\sqrt{3}}{4} - \left(\frac{\pi}{6} + \frac{\sqrt{3}}{4}\right)\right) = \frac{25\pi}{3}\).
**Notes:**
- The solution involves understanding and applying trigonometric identities and substitution techniques.
- Highlights the use of integral boundaries transformation in trigonometric integrals.
This solution demonstrates a clear application of trigonometric substitution in solving definite integrals, a crucial topic in Calculus II studies.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2F28f649eb-2553-4502-ac79-c1e09b799c37%2Fp74iyq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The text is a mathematical solution to a definite integral problem, involving trigonometric substitution. It is appropriate for an educational website explaining calculus concepts.
**Problem Description:**
- **Given Integral:** \(\int_{5}^{5\sqrt{3}} \sqrt{100 - x^2} \, dx\)
**Solution Steps:**
1. **Substitution:**
- Let \(x = 10 \sin \theta\), which implies \(dx = 10 \cos \theta \, d\theta\).
- Notice that \(\sqrt{100 - x^2} = 10 \cos \theta\).
2. **Changing Limits of Integration:**
- When \(x = 5\), \(\theta = \frac{\pi}{6}\).
- When \(x = 5\sqrt{3}\), \(\theta = \frac{\pi}{3}\).
3. **Integral Transformation:**
- The integral becomes \(\int_{\pi/6}^{\pi/3} 100 \cos^2 \theta \, d\theta\).
4. **Use of Trigonometric Identity:**
- Convert \(\cos^2 \theta\) using the identity \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\).
5. **Integral Evaluation:**
- Compute \(\int_{\pi/6}^{\pi/3} 50 (1 + \cos 2\theta) \, d\theta\).
- The result simplifies to \(50 \left[ \theta + \frac{\sin 2\theta}{2} \right]_{\pi/6}^{\pi/3}\).
6. **Final Calculation:**
- Perform the final computation to get \(50 \left(\frac{\pi}{3} + \frac{\sqrt{3}}{4} - \left(\frac{\pi}{6} + \frac{\sqrt{3}}{4}\right)\right) = \frac{25\pi}{3}\).
**Notes:**
- The solution involves understanding and applying trigonometric identities and substitution techniques.
- Highlights the use of integral boundaries transformation in trigonometric integrals.
This solution demonstrates a clear application of trigonometric substitution in solving definite integrals, a crucial topic in Calculus II studies.
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