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
Surface area about x axis refers to the total area of a three-dimensional object when rotated around the x axis. This measurement helps to determine the amount of material needed to cover the surface of the object.
The formula for calculating surface area about x axis is 2π∫(f(x)√(1+(f'(x))^2)dx, where f(x) represents the function of the object's cross-section in terms of x. This integral can be solved using calculus.
Surface area about x axis is an important measurement in fields such as engineering, architecture, and manufacturing. It helps in determining the amount of material needed to construct an object, as well as in optimizing designs for efficiency and cost-effectiveness.
Surface area about x axis is specific to the rotation of an object around the x axis, while surface area about y and z axes refer to rotations around the y and z axes, respectively. This means that the cross-sections of the object and the formula for calculating surface area will be different for each axis.
No, surface area about x axis cannot be negative. It is always a positive value since it represents the total area of the object's surface, regardless of its orientation.