Answer to Question #193254 in Calculus for April

Question

Determine the surface area generated by revolving the arc 3y= 4x from x=0 to x=3 about the x-axis.

Answer 1

3y = 4x

Surface Area generated by revolving the arc 3y= 4x from x=0 to x=3 about the x-axis is given by

Area = ∫₀³ 2 $\pi$ * y * $\sqrt{\smash[b]{1 + (dy/dx)^2 }}$ dx ..............Equation(1)

dy/dx = 4/3 = 1.33

Area = ∫₀³ 2 $\pi$ * 1.33x * $\sqrt{1 + 1.33^2}$ dx

Area = ∫₀³ 2 $\pi$* 1.33x * 1.664 dx

Area = 13.89 ∫₀³ x dx

Area = 13.89 * ( x² / 2) ₀³

Area = 13.89 * 4.5

Area = 62.54 square units