Mass double integral?
Let S be a thin surface with uniform thickness of .01 m having the shape of the part of the paraboloid z=x^2+y^2 below the plane z=1. Suppose the surface density at each point P(x,y,z) is a constant 2kg/m3. Find the mass of the surface
I tried taking the volume and multiplying it by the density but that didn't get me the right answer. Any help is appreciated
Comments
A = ∫ ∫ √((2x)² + (2y)² + 1) dx dy .... is the surface area of the shell
... let x = r cos(θ) and y = r sin(θ)
... ∂(r,θ) / ∂(x,y) = r
A = ∫ ∫ r √(4r² + 1) dr dθ [0,1] [0,2π]
V ≈ 1/100 ∫ ∫ r √(4r² + 1) dr dθ [0,1] [0,2π] ... since the shell is thin
M ≈ 1/50 ∫ ∫ r √(4r² + 1) dr dθ [0,1] [0,2π]
= π/25 ∫ r √(4r² + 1) dr [0,1]
= π/300 (5√5 - 1)
Answer: π/300 (5√5 - 1) kg ... ≈ 0.107 kg
word that the area is bounded is merely in the 1st quadrant hence your area for this area is D {(x,y)| 0 ? x ? a million, 0 ? y ? a million} observe that all of them are 0 and a million, it somewhat is by using the fact we basically care pertaining to to the 1st quadrant, the place the two curves basically intersect.