how to find double integral to polar coordinate

Question

₀∫² _y∫^{√8-y^2} √X^2+y^2 dxdy

Answer 1

The question asks us to convert this integral to polar coordinate:

∫(y=0->2) ∫(x=y->sqrt(8 - y^2)) sqrt(x^2 + y^2) dx dy

To do this, we can use the typical conversions from cartesian to polar coordinates:

1. x = r*cos(θ)
2. y = r*sin(θ)
3. x^2 + y^2 = r^2
4. dA = dx*dy = r dr dθ

Calculating new bounds:

1. Inside the circle x^2 + y^2 = 8 → r^2 = 8 → r = sqrt(8)
2. Above y = 0 → first quadrant
3. To the right of x = y → that line is θ = pi/4

So in polar:

1. 0 <= θ <= pi/4
2. 0 <= r <= sqrt(8)

Let's convert the integrand now:

sqrt(x^2 + y^2) = sqrt((r^2)*sin^2+cos^2) = sqrt((r^2)*1) = r

dx dy = r dr dθ

Plugging in:

sqrt(x^2 + y^2) * dx * dy

= r * (r dr dθ)

= r^2 dr dθ

Now let's put everything together:

∫(y=0->2) ∫(x=y->sqrt(8 - y^2)) sqrt(x^2 + y^2) dx dy

becomes

∫(θ=0->pi/4) ∫(r=0->sqrt(8)) r^2 dr dθ