Answer:
[tex]\displaystyle \int\limits^{\frac{\pi}{6}}_0 \int\limits^{2 \sec \theta}_0 {r^5 \sin 2\theta} \, dr \, d\theta = \int\limits^{\frac{\sqrt{3}}{3}x}_0 \int\limits^2_0 {2xy(x^2 + y^2)} \, dx \, dy[/tex]
General Formulas and Concepts:
Pre-Calculus
Polar Conversions:
- [tex]\displaystyle x = r \cos \theta[/tex]
- [tex]\displaystyle y = r \sin \theta[/tex]
- [tex]\displaystyle \tan \theta = \frac{y}{x}[/tex]
Multivariable Calculus
Double Integrals
Integral Conversion [Polar Coordinates]:
[tex]\displaystyle \iint_R {F(r, \theta)} \, dA = \iint_R {F(r, \theta)r} \, dr \, d\theta[/tex]
Step-by-step explanation:
Step 1: Define
Identify given.
[tex]\displaystyle \int\limits^{\frac{\pi}{6}}_0 \int\limits^{2 \sec \theta}_0 {r^5 \sin 2\theta} \, dr \, d\theta[/tex]
Step 2: Convert Pt. 1
Find y bounds.
- [Polar Conversions] Substitute in θ [dθ bound]:
[tex]\displaystyle \tan \frac{\pi}{6} = \frac{y}{x}[/tex] - Evaluate [Unit Circle]:
[tex]\displaystyle \frac{\sqrt{3}}{3} = \frac{y}{x}[/tex] - Rewrite:
[tex]\displaystyle y = \frac{\sqrt{3}}{3}x[/tex] - Define limits:
[tex]\displaystyle 0 \leq y \leq \frac{\sqrt{3}}{3}x[/tex]
Find x bounds.
- [dr bound] Set up:
[tex]\displaystyle 2 \sec \theta = r[/tex] - Rewrite:
[tex]\displaystyle r \cos \theta = 2[/tex] - Substitute in Polar Conversions:
[tex]\displaystyle x = 2[/tex] - Define limits:
[tex]\displaystyle 0 \leq x \leq 2[/tex]
Step 3: Convert Pt. 2
- [Integrals] Rewrite integrand:
[tex]\displaystyle \int\limits^{\frac{\pi}{6}}_0 \int\limits^{2 \sec \theta}_0 {r^5 \sin 2\theta} \, dr \, d\theta = \int\limits^{\frac{\pi}{6}}_0 \int\limits^{2 \sec \theta}_0 {\big[ 2r^2 r \sin (\theta) r \cos (\theta) \big] r} \, dr \, d\theta[/tex] - [Integrals] Rewrite [Integral Conversion [Polar Coordinates]:
[tex]\displaystyle \int\limits^{\frac{\pi}{6}}_0 \int\limits^{2 \sec \theta}_0 {r^5 \sin 2\theta} \, dr \, d\theta = \int\limits^{\frac{\pi}{6}}_0 \int\limits^{2 \sec \theta}_0 {2r^2 r \sin (\theta) r \cos (\theta)} \, dA[/tex] - [Integrand] Substitute in Polar Conversions:
[tex]\displaystyle \int\limits^{\frac{\pi}{6}}_0 \int\limits^{2 \sec \theta}_0 {r^5 \sin 2\theta} \, dr \, d\theta = \int\limits^{\frac{\pi}{6}}_0 \int\limits^{2 \sec \theta}_0 {2xy(x^2 + y^2)} \, dA[/tex] - [Bounds] Rewrite:
[tex]\displaystyle \int\limits^{\frac{\pi}{6}}_0 \int\limits^{2 \sec \theta}_0 {r^5 \sin 2\theta} \, dr \, d\theta = \int\limits^{\frac{\sqrt{3}}{3}x}_0 \int\limits^2_0 {2xy(x^2 + y^2)} \, dx \, dy[/tex]
∴ we have converted the polar integral into a rectangular integral.
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Learn more about double integrals: https://brainly.com/question/17433118
Learn more about multivariable calculus: https://brainly.com/question/19234614
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Topic: Multivariable Calculus
Unit: Double Integrals
