Math Problem Statement

Evaluate the integral ∫ (from 0 to 2) r^2 * sqrt(4 - r^2) dr

Solution

The given integral is:

∫02r24−r2 dr\int_0^2 r^2 \sqrt{4 - r^2} \, dr

Solution:

We can solve this by using a trigonometric substitution. Let’s use the substitution:

r=2sin⁡θr = 2 \sin \theta

This implies: dr=2cos⁡θ dθdr = 2 \cos \theta \, d\theta and r2=4sin⁡2θr^2 = 4 \sin^2 \theta

Substituting r=2sin⁡θr = 2 \sin \theta into the integral, the limits of integration will change as follows:

  • When r=0r = 0, θ=0\theta = 0
  • When r=2r = 2, θ=π2\theta = \frac{\pi}{2}

Thus, the integral becomes:

∫0π2(4sin⁡2θ)⋅(2cos⁡θ)⋅2cos⁡θ dθ\int_0^{\frac{\pi}{2}} (4 \sin^2 \theta) \cdot (2 \cos \theta) \cdot 2 \cos \theta \, d\theta

This simplifies to:

16∫0π2sin⁡2θcos⁡2θ dθ16 \int_0^{\frac{\pi}{2}} \sin^2 \theta \cos^2 \theta \, d\theta

We can now use the double-angle identity for sin⁡2θ\sin^2 \theta and cos⁡2θ\cos^2 \theta:

\quad \text{and} \quad \cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$$ Now substitute these identities into the integral: $$16 \int_0^{\frac{\pi}{2}} \frac{(1 - \cos(2\theta))(1 + \cos(2\theta))}{4} \, d\theta$$ This simplifies to: $$4 \int_0^{\frac{\pi}{2}} (1 - \cos^2 (2\theta)) \, d\theta$$ Now, simplify the expression: $$4 \left[ \int_0^{\frac{\pi}{2}} 1 \, d\theta - \int_0^{\frac{\pi}{2}} \cos^2 (2\theta) \, d\theta \right]$$ The first integral is straightforward: $$\int_0^{\frac{\pi}{2}} 1 \, d\theta = \frac{\pi}{2}$$ For the second integral, we use the identity: $$\cos^2 (x) = \frac{1 + \cos (2x)}{2}$$ Applying this, we get: $$\int_0^{\frac{\pi}{2}} \cos^2 (2\theta) \, d\theta = \frac{1}{2} \int_0^{\frac{\pi}{2}} \left(1 + \cos (4\theta)\right) \, d\theta$$ The first part is simple: $$\frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$$ The second part is: $$\frac{1}{2} \int_0^{\frac{\pi}{2}} \cos (4\theta) \, d\theta = 0$$ because the integral of $$\cos (4\theta)$$ over $$[0, \frac{\pi}{2}]$$ is zero. Thus, the second integral is $$\frac{\pi}{4}$$, and the full expression becomes: $$4 \left( \frac{\pi}{2} - \frac{\pi}{4} \right) = 4 \cdot \frac{\pi}{4} = \pi$$ ### Final Answer: $$\boxed{\pi}$$ Do you want more details or have any questions? Here are some relative questions to explore: 1. How does trigonometric substitution work in integrals? 2. Why is $$r = 2 \sin \theta$$ a suitable substitution for this problem? 3. What is the double-angle identity and how is it useful? 4. What are some other methods for evaluating integrals involving square roots? 5. How does the integral of $$\cos(k\theta)$$ behave over standard intervals? **Tip:** When dealing with integrals involving square roots of quadratic expressions, trigonometric substitution often simplifies the expression significantly.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometric Substitution
Integral Calculus
Double Angle Identity

Formulas

r = 2sin(θ), dr = 2cos(θ)dθ
sin^2(θ) = (1 - cos(2θ))/2
cos^2(θ) = (1 + cos(2θ))/2

Theorems

Double Angle Identity
Integral of cos(kθ)

Suitable Grade Level

Undergraduate - Calculus II

Related Recommendation

Evaluating the Integral ∫_0^1 √(1 - x^2) dx Using Trigonometric Substitution

Evaluate the Integral of √(r² - x²) from 0 to r

Evaluate \(\int_{0}^{2}\int \sqrt{4-x^2}\,dx\,dx\): Step-by-Step Solution

Evaluate Integral with Trigonometric Substitution: \(x = 4\sin^2\theta\)

Evaluate the Integral ∫ (x / √(4 - x^2)) dx Using Trigonometric Substitution