The standard formula for the curved surface of revolution about the x-axis for a continuous function f(x) is:
A=∫2πf(x)sqrt(1+(f'(x))²)dx
Substitute values
f(x)=sqrt(1+5x)
f'(x)=5/(2sqrt(5x+1))
integrand:
2πsqrt(5x+1)*sqrt(1+25/(4(5x+1)))
Limits: 0 to 7
Integral:
(π(20x+29)^(3/2))/30
Evaluate between 0 and 7 to get:
214 units² approximately.
Find the exact area of the surface obtained by rotating the curve about the x-axis.
y = sqrt(1 + 5x) from 1 ≤ x ≤ 7
6 answers
Think of the surface as a collection of strips, of radius y, and width a piece of the curve, ds
A = Int (2π r ds)
but ds = √(dx^2 + dy^2)
= √(1 + y'^2) dx
So, here we have
y = √(5x+1)
y' = 5/√(5x+1)
A = 2π Int(y ds)
= 2π Int(√(5x+1)*√(1+25/(5x+1)) dx[1,7]
= 2π Int(√(5x+26))[1,7]
= 2π * 2/3 * 1/5 (5x+26)^3/2[1,7]
= 4π/15 (61√61 - 31√31)
A = Int (2π r ds)
but ds = √(dx^2 + dy^2)
= √(1 + y'^2) dx
So, here we have
y = √(5x+1)
y' = 5/√(5x+1)
A = 2π Int(y ds)
= 2π Int(√(5x+1)*√(1+25/(5x+1)) dx[1,7]
= 2π Int(√(5x+26))[1,7]
= 2π * 2/3 * 1/5 (5x+26)^3/2[1,7]
= 4π/15 (61√61 - 31√31)
Rats. MathMate did it right.
y' = 5/2√(5x+1)
Sorry
y' = 5/2√(5x+1)
Sorry
Except for the limits of integration :-)
Good point!
I'll let Michael take care of that!
Michael:
Limits should be from 1 to 7.
I'll let Michael take care of that!
Michael:
Limits should be from 1 to 7.
The answer is 309pi/5 with the limit from 1 to 7
1 answer