∫ 0 6 ​ ∫ 0 y ​ xdxdy convert the equation into polar form

To convert the given double integral \(\int_0^6 \int_0^y x \, dx \, dy\) into polar coordinates, we need to express both the region of integration and the integrand in terms of the polar variables \(r\) (radius) and \(\theta\) (angle).

### Steps for Conversion:

1. **Conversion of Coordinates:**
   In polar coordinates:
   \[
   x = r \cos(\theta)
   \]
   \[
   y = r \sin(\theta)
   \]
   The Jacobian determinant of the transformation from Cartesian to polar coordinates is \(r\), so when converting to polar form, we need to include this factor in the integrand.

2. **Bounds of Integration:**
   - In the given Cartesian bounds, \(x\) goes from \(0\) to \(y\), and \(y\) goes from \(0\) to \(6\).
   - In polar coordinates, \(r\) represents the radial distance, and \(\theta\) represents the angle.
     - \(y = r \sin(\theta)\), which is non-negative for \(\theta \in [0, \frac{\pi}{2}]\).
     - For \(y\) ranging from \(0\) to \(6\), \(r\) will range from \(0\) to \(6\), and \(\theta\) will range from \(0\) to \(\frac{\pi}{2}\) (since we're in the first quadrant).

3. **Rewrite the Integral:**
   The integral becomes:
   \[
   \int_0^{\frac{\pi}{2}} \int_0^6 r^2 \cos(\theta) \, dr \, d\theta
   \]
   Here, \(r^2 \cos(\theta)\) comes from \(x = r \cos(\theta)\), and the factor of \(r\) comes from the Jacobian determinant.

### Solving the Integral:

Now, let's break down the integral:
\[
\int_0^{\frac{\pi}{2}} \int_0^6 r^2 \cos(\theta) \, dr \, d\theta
\]

1. **Integrate with respect to \(r\):**
   \[
   \int_0^6 r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^6 = \frac{6^3}{3} = 72
   \]

2. **Now, integrate with respect to \(\theta\):**
   \[
   72 \int_0^{\frac{\pi}{2}} \cos(\theta) \, d\theta = 72 \left[ \sin(\theta) \right]_0^{\frac{\pi}{2}} = 72 \times 1 = 72
   \]

### Final Answer:
The value of the double integral in polar form is:
\[
72
\]

---

Would you like more details or clarifications on any part of this process? 

Here are five related questions to consider:
1. How do you find the Jacobian determinant when changing to polar coordinates?
2. Why do we multiply by \(r\) when converting to polar coordinates?
3. What are the general steps to convert an integral from Cartesian to polar form?
4. How do we determine the bounds of integration in polar coordinates?
5. How do trigonometric functions like \(\cos(\theta)\) and \(\sin(\theta)\) relate to \(x\) and \(y\) in polar coordinates?

**Tip:** When converting to polar coordinates, always remember to include the Jacobian factor \(r\), which accounts for the area element in polar coordinates.

Explore the step-by-step process of converting the double integral ∫ 0 6 ∫ 0 y xdxdy from Cartesian coordinates to polar form. Learn key concepts like polar coordinate transformations, the Jacobian determinant, and solving the resulting integral. Suitable for students studying undergraduate calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation