The discussion focuses on calculating the surface area of revolution about the x-axis for the equation x² + (y - 2)² = 1. The user successfully derived the functions y = √(1 - x²) + 2 and y = 2 - √(1 - x²) but encountered difficulties with integration limits, specifically when y = 0, leading to complex calculations. The formula used for surface area is ∫ 2π * (√(1 - x²) + 2)√(1 + x²/(1 - x²))dx, which requires careful consideration of the curve's graph to determine appropriate limits.
PREREQUISITESStudents in calculus courses, mathematics educators, and anyone interested in advanced integration techniques and geometric applications in calculus.
That just means that y=0 does not happen. Your integration limits are determined by something else...mikbear said:when y= 0; x = - √(-3) and x = √(-3)
How did you get that formula?I use the formula ∫ 2∏ * (√(1-x^2) + 2)√(1 + x^2/(1 - x^2))dx
mikbear said:Homework Statement
x^2 + (y - 2)^2 = 1
The hint given by the question was to split the function into 2
Homework Equations
Surface Area about x axis
The Attempt at a Solution
So i did this.
(y - 2)^2 = 1 - x^2
y = √(1 - x^2) +2 and y = 2 - √(1 - x^2)
The range I calculated
when y= 0; x = - √(-3) and x = √(-3)
I use the formula ∫ 2∏ * (√(1-x^2) + 2)√(1 + x^2/(1 - x^2))dx
Then i got stuck as it became messy