# How do you find the area of the surface generated by rotating the curve about the y-axis #y=cx+d, a<=x<=b#?

##### 2 Answers

The surface area for

#S = pi(b^2 - a^2)sqrt(c^2 + 1)# ,#" "" "c > 0#

(Note: If

If

#y_(min) = ca + d#

#y_(max) = cb + d# (If

#c < 0# , then the revolution around the#y# axis would produce the same surface area by symmetry of#y = |c|x + d# compared to#y = -|c|x + d# . The value of#d# does not matter, since the rotation is around the#y# axis, and the shape will be the same regardless of the y-intercept.)

The **surface area** for a revolution around the

#S = 2 pi int_(ca + d)^(cb + d) f(y) sqrt(1 + ((dx)/(dy))^2)dy#

Evaluating the derivative, we get:

#((dx)/(dy))^2 = (d/(dy)[(y - d)/c])^2#

#= 1/c^2#

Therefore:

#S = 2 pi int_(ca + d)^(cb + d) (y - d)/c sqrt(1 + 1/c^2)dy#

#= (2 pi sqrt(1 + 1/c^2))/c int_(ca + d)^(cb + d) (y - d) dy#

#= (2 pi sqrt(1 + 1/c^2))/c |[y^2/2 - dcdoty]|_(ca + d)^(cb + d)#

#= (2 pi sqrt(1 + 1/c^2))/c [(cb + d)^2/2 - dcdot(cb + d)] - [(ca + d)^2/2 - dcdot(ca + d)]#

#= (2 pi sqrt(1 + 1/c^2))/c [(cb + d)^2/2 - dcdot(cb + d) - (ca + d)^2/2 + dcdot(ca + d)]#

#= (2 pi sqrt(1 + 1/c^2))/c [(b^2c^2 + 2bcd + d^2)/2 - bcd - d^2 - (a^2c^2 + 2acd + d^2)/2 + acd + d^2]#

#= (2 pi sqrt(1 + 1/c^2))/c [(b^2c^2 + 2bcd + cancel(d^2) - a^2c^2 - 2acd - cancel(d^2))/2 - bcd - cancel(d^2) + acd + cancel(d^2)]#

#= (2 pi sqrt(1 + 1/c^2))/cancel(c) [(cancel(c)(b^2c + 2d(b - a) - a^2c))/2 + cancel(c)d(a - b)]#

#= 2 pi sqrt(1 + 1/c^2) [(c(b^2 - a^2) + 2d(b - a))/2 + (2d(a - b))/2]#

#= 2 pi sqrt(1 + 1/c^2) [(c(b^2 - a^2))/2]#

#= 2 pi sqrt(c^2 + 1) cdot (b^2 - a^2)/2#

#= bb(color(blue)(pi (b^2 - a^2)sqrt(c^2 + 1)))#

For example, let's rotate

#S stackrel(?" ")(=) pi (2^2 - 0^2)sqrt(2^2 + 1)#

#stackrel(?" ")(=) 4sqrt5pi#

In Wolfram Alpha, we get that it works!

# A = pi(b^2-a^2)sqrt(1+c^2) #

#### Explanation:

The function:

# y=cx+d#

represents a straight line. Rotating a straight line about

# A= pi(R+r)sqrt((R-r)^2+h^2) #

Where:

#R# is the lower radius

#r# is the upper radius

#h# is the vertical height

When:

# x=a => y=ca+d #

# x=b => y=cb+d #

Thus we have:

# r=a, R=b #

# h=(cb+d)-(ca+d) = cb-ca #

Thus, the SA, is:

# A = pi(a+b)sqrt((b-a)^2+(cb-ca)^2) #

# \ \ \ = pi(a+b)sqrt((b-a)^2+c^2(b-a)^2) #

# \ \ \ = pi(a+b)sqrt((b-a)^2(1+c^2)) #

# \ \ \ = pi(a+b)(b-a)sqrt(1+c^2) #

# \ \ \ = pi(b^2-a^2)sqrt(1+c^2) #