Find the exact area of the surface obtained by rotating the curve about the x-axis.
y = sqrt(1+5x) from 1<x<7
Integral of sqrt(1+5x) = 2/15 (1+5x)^3/2
evaluate at 7 and 1, and we have
4/5 (36-sqrt(6))
Oops. my bad. I read area, not surface area. MathMate did it right.
To find the exact area of the surface obtained by rotating the curve about the x-axis, we can use the method of "disk integration." This involves integrating the areas of infinitesimally thin disks that make up the surface.
First, let's express the equation of the curve in terms of x: y = √(1 + 5x).
To find the infinitesimal area of a disk, we need to consider a small element of the curve (dy) and integrate it over the range of x-values given (1 to 7).
The formula for the infinitesimal area of a disk, dA, is given by: dA = 2πx(y) dy, where x(y) is the inverse function of y(x).
To find the inverse function of y(x), we can square both sides of the original equation: y^2 = 1 + 5x. Rearranging, we get: x = (y^2 - 1)/5.
Now, we need to find the limits of integration for y. With 1 < x < 7, we can substitute the x-values into the inverse function to find the corresponding y-values: y = √(1 + 5x).
When x = 1, y = √(1 + 5(1)) = √6.
When x = 7, y = √(1 + 5(7)) = √36 = 6.
So, the limits of integration for y are from √6 to 6.
Now, we can set up the integral for the total surface area:
A = ∫[√6, 6] 2πx(y) dy.
Substituting x = (y^2 - 1)/5, we get:
A = ∫[√6, 6] 2π((y^2 - 1)/5)(y) dy.
Now, we can integrate this expression:
A = 2π/5 ∫[√6, 6] (y^3 - y) dy.
Evaluating this integral, we get:
A = 2π/5 [(1/4)y^4 - (1/2)y^2] | [√6, 6].
Calculating the values at the limits of integration:
A = 2π/5 [((1/4)(6^4) - (1/2)(6^2)) - ((1/4)(√6^4) - (1/2)(√6^2))].
Simplifying this expression will give us the exact area of the surface obtained by rotating the curve about the x-axis.