Question

Find the area of the surface generated by revolving the curve about the axis y = 9-x^2

Answer 1

The area of the surface generated by revolving the curve about the axis y = 9-x² is 12π

What is a revolution on a graph? or curve?

Surfaces of revolution are graphs of functions f(x,y) that rely exclusively on the distance from the origin of the point (x,y).

Polar coordinates (r,) are one approach to discuss such surfaces. A "surface of revolution" is formed by rotating a curve around an axis.

Consider a curve with coordinates x and z in the "x0" half of the Euclidean plane.

What is the calculation justifying the above answer?

y = √(9-x²)

⇒ dy = (-2x)/2√9-x²)

= -x/√(9-x²)

Note that surface area (s) is given as:

$\int\limits^b_a {2\pi y}$ √1 + (dy/dx)² dx

= 2$\pi$ $\int_{1}^{-1}$ √9-x² * √[1 + (-x)/√9-x²)]dx

= 2π $\int_{1}^{-1}$3dx
= 6π [x]₋¹₁

= 6π (1-(-1)

= 6π (2)

= 12π

Learn more about revolving on a curve:
brainly.com/question/2293431
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