Find the exact area of the surface obtained by rotating the curve about the x-axis.
y = sqrt(1 + 5x) from 1 ≤ x ≤ 7
The answer is 309pi/5 with the limit from 1 to 7
Good point!
I'll let Michael take care of that!
Michael:
Limits should be from 1 to 7.
Rats. MathMate did it right.
y' = 5/2√(5x+1)
Sorry
Except for the limits of integration :-)
To find the exact area of the surface obtained by rotating the curve about the x-axis, we need to use the formula for the surface area of a solid of revolution.
The formula for the surface area of a solid of revolution is given by:
S = ∫2πy * ds
where S is the surface area, y is the height of the curve at a particular x-value, and ds is an element of arc length.
To find ds, we can use the formula:
ds = √(1 + (dy/dx)²) * dx
In this case, we have the equation of the curve as y = √(1 + 5x), so we can find dy/dx by taking the derivative of y with respect to x.
dy/dx = (d/dx)√(1 + 5x) = (5/2)√(1 + 5x)
Now, we can substitute dy/dx into the formula for ds:
ds = √(1 + (dy/dx)²) * dx
= √(1 + ((5/2)√(1 + 5x))²) * dx
Substituting back into the formula for the surface area:
S = ∫2πy * √(1 + ((5/2)√(1 + 5x))²) * dx
Now, we can integrate this expression with respect to x, from x = 1 to x = 7, to find the exact area of the surface.
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)
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.