How do you find the surface area generated when the curve #y=sqrt(r^2-x^2), 0<=x<=a# is rotated about the x axis?

2 Answers
Apr 17, 2018

Take a piece of the curve whose length is the differential of arc length #ds = sqrt(1+(f'(x))^2) dx#

The surface area of the corresponding piece of the solid is

the circumference of revolution time the differential of arc length

which is

#2pif(x)sqrt(1+(f'(x))^2) dx#

Using just the values #0 <= x <=a#, the surface area of the solid is

#2piint_0^a f(x)sqrt(1+(f'(x))^2) dx = 2pi int_0^a sqrt(r^2-x^2)sqrt(1+((-x)/sqrt(r^2-x^2))^2) dx#

# = 2pi int_0^a r dx#

# = 2pi r a#

Apr 17, 2018

#A=2piar#

Explanation:

The formular is
#A=2piint_a^bf(x)sqrt(1+(f'(x))^2)dx#
#f(x)=y=sqrt(r^2-x^2)#
#f'(x)=-(x)/(sqrt(r^2-x^2)#
#A=2piint_0^asqrt(r^2-x^2)sqrt(1+(x^2)/(r^2-x^2))dx#
#<=>2piint_0^asqrt(r^2-x^2)sqrt(r^2/(r^2-x^2))dx#
#<=>2piint_0^ardx#
#<=>2pi[xr]_0^a#
#<=>2piar#