Math Problem Statement

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

Solution

To convert the given double integral 060yxdxdy\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 rr (radius) and θ\theta (angle).

Steps for Conversion:

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

  2. Bounds of Integration:

    • In the given Cartesian bounds, xx goes from 00 to yy, and yy goes from 00 to 66.
    • In polar coordinates, rr represents the radial distance, and θ\theta represents the angle.
      • y=rsin(θ)y = r \sin(\theta), which is non-negative for θ[0,π2]\theta \in [0, \frac{\pi}{2}].
      • For yy ranging from 00 to 66, rr will range from 00 to 66, and θ\theta will range from 00 to π2\frac{\pi}{2} (since we're in the first quadrant).
  3. Rewrite the Integral: The integral becomes: 0π206r2cos(θ)drdθ\int_0^{\frac{\pi}{2}} \int_0^6 r^2 \cos(\theta) \, dr \, d\theta Here, r2cos(θ)r^2 \cos(\theta) comes from x=rcos(θ)x = r \cos(\theta), and the factor of rr comes from the Jacobian determinant.

Solving the Integral:

Now, let's break down the integral: 0π206r2cos(θ)drdθ\int_0^{\frac{\pi}{2}} \int_0^6 r^2 \cos(\theta) \, dr \, d\theta

  1. Integrate with respect to rr: 06r2dr=[r33]06=633=72\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 θ\theta: 720π2cos(θ)dθ=72[sin(θ)]0π2=72×1=7272 \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: 7272


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 rr 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(θ)\cos(\theta) and sin(θ)\sin(\theta) relate to xx and yy in polar coordinates?

Tip: When converting to polar coordinates, always remember to include the Jacobian factor rr, which accounts for the area element in polar coordinates.

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Math Problem Analysis

Mathematical Concepts

Double Integrals
Polar Coordinates
Coordinate Transformations

Formulas

x = r cos(θ)
y = r sin(θ)
Jacobian determinant for polar coordinates: r

Theorems

Change of Variables Theorem
Jacobian Determinant in Coordinate Transformations

Suitable Grade Level

Undergraduate Calculus (Grades 11-12 or early university level)