Evaluate the double integral ∬Dx6ydA, where D is the top half of the disc with center the origin and radius 7, by changing to polar coordinates.

Answer 1

Consider the double integral ∫∫_D x⁶y dA, where the region D is the top half of the disc (circle) with radius 7 centered at the origin.

The relation between rectangular (x,y) and polar (r,θ) coordinates is:

r² = x² + y² , x = rcosθ, and y = rsinθ

Region D is a semi-circle ("top half of the disc"), so we can define the region D in polar coordinates as:

0 ≤ r ≤ 7 and 0 ≤ θ ≤ π, therefore D = {(r,θ) | 0 ≤ r ≤ 7, 0 ≤ θ ≤ π}

To evaluate the double integral, we start by substituting x = rcosθ and y = rsinθ into the original equation:

∫∫_D x⁶y dA = ∫₀^π∫₀⁷_D (rcosθ)⁶ rsinθ rdrdθ

[Remember: must multiply by that extra r when converting double integrals into polar coordinates]

= ∫₀^π∫₀⁷_D (r⁶cos⁶θ) rsinθ rdrdθ

= ∫₀^π∫₀⁷_D r⁸cos⁶θ sinθ drdθ

= ∫0π cos⁶θ sinθ [r⁹/9]₀⁷dθ

= (7⁹/9) ∫₀^π cos⁶θ sinθ dθ

To integrate the next integral, use u-substitution where u = cosθ, du = -sinθ dθ:

= (7⁹/9) ∫ u⁶ (sinθ/-sinθ) du

= - (7⁹/9) [u⁷/7] [The negative sign comes from (sinθ/-sinθ) = -1]

= - (7⁹/9) [cos⁷θ/7]₀^π

= - (7⁹/63) [cos⁷(π) - cos⁷(0)] [Pull the 7 in the denominator out front]

= - (7⁹/63) [(-1)⁷ - (1)⁷]

= - (7⁹/63) [(-1) - (1)]

= -(7⁹/63) × (-2)

= 11529602 / 9

≈ 1281066.889

Not a pretty result but should be correct. Hope this helps!