Evaluate Double Integrals Using Polar Coordinates

Math Problem Statement

Integrate the given double integrals by converting them to polar coordinates.

This is a problem that requires converting Cartesian double integrals to polar coordinates for evaluation. Let's solve both integrals step by step:

$\int_0^a \int_{-\sqrt{a^2 - y^2}}^{\sqrt{a^2 - y^2}} x^2 \, dx \, dy$

The $x$ -bound is from $-\sqrt{a^2 - y^2}$ to $\sqrt{a^2 - y^2}$ , indicating a semicircle in the $x$ -direction for a fixed $y$ .
The $y$ -bound is from $0$ to $a$ , representing the top half of a circle of radius $a$ centered at the origin.

In polar coordinates:

The region corresponds to $r \in [0, a]$ and $\theta \in [0, \pi]$ .

$x^2 \, dx \, dy = (r\cos\theta)^2 r \, dr \, d\theta$

The integral becomes:

= \int_0^\pi \cos^2\theta \, d\theta \int_0^a r^3 \, dr$$ #### **Step 3: Evaluate the integrals** 1. $$\int_0^a r^3 \, dr = \left[\frac{r^4}{4}\right]_0^a = \frac{a^4}{4}$$. 2. $$\int_0^\pi \cos^2\theta \, d\theta = \int_0^\pi \frac{1 + \cos(2\theta)}{2} \, d\theta = \frac{1}{2} \left[\theta + \frac{\sin(2\theta)}{2}\right]_0^\pi = \frac{\pi}{2}$$. Thus, the result is: $$\frac{a^4}{4} \cdot \frac{\pi}{2} = \frac{\pi a^4}{8}.$$ --- ### **2. Problem 2** $$\int_0^a \int_{\sqrt{x-x^2}}^{\sqrt{x^2 + y^2}} \sqrt{x^2 + y^2} \, dy \, dx$$ #### **Step 1: Analyze the bounds** - The $$y$$-bound is from $$\sqrt{x - x^2}$$ to $$\sqrt{x^2 + y^2}$$, which appears ambiguous or complex. Likely, the limits refer to a circular region, and further clarification of bounds might be helpful. Could you clarify this part to ensure a detailed solution? --- Let me know if you'd like more details for **Problem 2** or refinements on **Problem 1**! **Relative Questions:** 1. What are the key steps in converting a double integral to polar coordinates? 2. How is the Jacobian determined in polar transformations? 3. How does the region of integration affect the bounds in polar coordinates? 4. What are the properties of $$\cos^2\theta$$ in integrals over symmetric intervals? 5. How do polar coordinates simplify integration over circular regions? **Tip:** When working with polar transformations, always sketch the region of integration to confirm bounds!

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Double Integrals
Polar Coordinates
Jacobian Transformation
Trigonometric Integration

x = r * cos(theta), y = r * sin(theta)
x^2 + y^2 = r^2
Jacobian determinant: r in polar coordinates
cos^2(theta) = (1 + cos(2theta))/2

Conversion of Cartesian integrals to polar coordinates
Symmetry in integration over circular regions

Undergraduate (Calculus II or III)

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