Find the volume of the solid under the surface z=(7+x)(13+y) and above the region in the xy-plane given by x^2+y^2<=36. Hint: Use polar coordinates
I tried to solve this question multiple times but i still cant seem to get the right answers i tried to enter 16699.9
I tried to solve this question multiple times but i still cant seem to get the right answers i tried to enter 16699.9
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V = ∫∫ (7+x)(13+y) dA
..= ∫(r = 0 to 6) ∫(θ = 0 to 2π) (7 + r cos θ)(13 + r sin θ) * (r dθ dr), via polar coordinates
..= ∫(r = 0 to 6) ∫(θ = 0 to 2π) (91r + 7r^2 sin θ + 13r^2 cos θ + r^3 sin θ cos θ) dθ dr
..= ∫(r = 0 to 6) (91rθ - 7r^2 cos θ + 13r^2 sin θ + (1/2) r^3 sin^2(θ)) {for θ = 0 to 2π} dr
..= ∫(r = 0 to 6) 182πr dr
..= 91πr^2 {for r = 0 to 6}
..= 3276π.
I hope this helps!
..= ∫(r = 0 to 6) ∫(θ = 0 to 2π) (7 + r cos θ)(13 + r sin θ) * (r dθ dr), via polar coordinates
..= ∫(r = 0 to 6) ∫(θ = 0 to 2π) (91r + 7r^2 sin θ + 13r^2 cos θ + r^3 sin θ cos θ) dθ dr
..= ∫(r = 0 to 6) (91rθ - 7r^2 cos θ + 13r^2 sin θ + (1/2) r^3 sin^2(θ)) {for θ = 0 to 2π} dr
..= ∫(r = 0 to 6) 182πr dr
..= 91πr^2 {for r = 0 to 6}
..= 3276π.
I hope this helps!