First we need to know how to calculate the lateral surface area of a conical frustum=πs(r₁+r₂) where s is the slant length and r₁ and r₂ are the radii of the upper and lower surfaces.
A conical frustum has straight sides, being part of a cone. Consider the given curve and we take two horizontal slices through it at y=a and y=b. We are interested in the line segment enclosed by these limits. If we now rotate the segment about the y-axis we get a frustum with a slant side s which is slightly curved (the line segment). When a and b are close together, s becomes more like a straight line so the rotated figure looks like a conical frustum. Call the slant side ds to denote an infinitesimal length. We can work out the x-coord for the y coords a and b. Call these aᵪ and bᵪ. If aᵪ=x then bᵪ=x+dx and the surface area of the frustum is πds(2x+dx). We can ignore dx in comparison to x, so this surface area, dA=2πxds.
We need to add up all such areas, which means A=∫2πxds. We can’t integrate this yet. We can relate ds to dy and dx: ds²=dy²+dx². We have x=4√(4-y), x=4(4-y)^½, so dx/dy=-4/(2√(4-y))=-2/√(4-y). ds/dy=√(1+(dx/dy)²), ds=√(1+(dx/dy)²)dy. So A=2πʃ4√(4-y)(√(1+4/(4-y))dy=8πʃ√(8-y)dy.
Let u²=8-y, then 2udu=-dy, so dy=-2udu and A=-8πʃu.2udu=-16πʃu²du=-(16π/3)u³.
At this point we can start thinking about limits for the definite integral. Instead of rewriting in terms of y, we can find limits for u. When lower limit y=0, u=√8=2√2. When upper limit y=55/16, u=√(8-55/16)=√(73/16).
When we substitute these values for u we get 215.84 sq units approx.