change the Cartesian integral into an equivalent polar integral

Question

\int _0^6\int _0^{y\:}xdxdy\:\:

Answer 1

We are given the Cartesian double integral:

∫ from 0 to 6 ∫ from 0 to y x dx dy

Our goal is to rewrite this integral using polar coordinates.

Step 1: Identify the region of integration.

The bounds tell us:

0 ≤ y ≤ 6

0 ≤ x ≤ y

This describes a triangular region in the first quadrant bounded by:

1. x = 0 (the y-axis)
2. x = y (a diagonal line)
3. y = 6 (a horizontal line)

Step 2: Convert the boundaries into polar coordinates.

Recall the polar coordinate formulas:

x = r cos(θ)

y = r sin(θ)

x² + y² = r²

Since the region lies between the positive y-axis and the line x = y, the angular bounds are:

π/4 ≤ θ ≤ π/2

Now consider the boundary y = 6.

Substitute y = r sin(θ):

r sin(θ) = 6

Solve for r:

r = 6/sin(θ)

The region starts at the origin, so:

0 ≤ r ≤ 6/sin(θ)

Step 3: Convert the integrand.

The original integrand is x.

Since x = r cos(θ), we substitute:

x = r cos(θ)

Remember that when converting a double integral to polar coordinates, the area element changes:

dx dy = r dr dθ

Therefore:

x dx dy = (r cos(θ))(r dr dθ)

= r² cos(θ) dr dθ

Step 4: Write the equivalent polar integral.

Putting everything together:

∫ from π/4 to π/2 ∫ from 0 to 6/sin(θ) r² cos(θ) dr dθ

Final Answer:

∫[π/4, π/2] ∫[0, 6/sin(θ)] r² cos(θ) dr dθ

This is the equivalent polar integral.

The most important things to remember are that the angle runs from π/4 to π/2, the upper radial bound comes from y = 6, and we must include the extra factor of r when converting dx dy to polar coordinates.