Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line, y = rad x, y = 0, x = 4, about x = 9

Answer 1

Using the method of cylindrical shells:

Draw a diagram.

Take a thin vertical cross section that is x units from the origin, has thickness Δx and has height √x, where x is between 0 and 4.

Rotating the cross section about the vertical line x = 9 yields a cylindrical shell with height √x, radius 9 - x, and thickness Δx.

dV = volume of shell = (circumference)(height)(thickness) = 2π(9-x) √xΔx = 2π(9x^1/2 - x^3/2)Δx

V = volume of solid = 2π∫_{(0 to 4)} [9x^1/2 - x^3/2]dx = 2π[6x^3/2 - (2/5)x^5/2]_{(0 to 4)} = 2π[6(8) - 64/5] = 352π/5

Answer 2

By "rad x", I assume you mean "radical x", meaning the square root function y = sqrt(x).

To write this integral, we need to integrate with respect to the y-axis, where our functions are now functions of y. To do this, simply solve for x in the original equations.

if y = sqrt(x), then x = y^2

y = 0 and x = 4 don't need to change.

x = y^2 intersects x = 4 when 4 = y^2, or when y = 2.

The circular slice of area will be the outer circular area minus the inner circular area.

dV = π(9 - y^2)^2 - π(9 - 4)^2 dy

Finally, the integral is just writing the formula for the slice of area together with the integral symbol and the bounds of integration.

V = ∫ dV = ∫₀² π(9 - y^2)^2 - π(9 - 4)^2 dy = π ∫₀² (9 - y^2)^2 - 25 dy