A) 15pi/16
B) 16pi
C) 2pi/3
D) 16pi/3
Thank you.
B) 16pi
C) 2pi/3
D) 16pi/3
Thank you.
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Since we are rotating the region about the x-axis, the surface area equals
∫ 2πy √(1 + (dy/dx)^2) dx
= ∫(x = 0 to 1) 2π * (3x/4) * √(1 + (3/4)^2) dx
= ∫(x = 0 to 1) 2π * (3x/4) * (5/4) dx
= (15π/16) * ∫(x = 0 to 1) 2x dx
= 15π/16.
I hope this helps!
∫ 2πy √(1 + (dy/dx)^2) dx
= ∫(x = 0 to 1) 2π * (3x/4) * √(1 + (3/4)^2) dx
= ∫(x = 0 to 1) 2π * (3x/4) * (5/4) dx
= (15π/16) * ∫(x = 0 to 1) 2x dx
= 15π/16.
I hope this helps!