At points on the curve y=2sqr(x), line segments of length h=y are drawn perpendicular to the xy-plane. Find the area of the surface formed by these perpendicular from (0.0) to (3.2sqr3).
To find the area of the surface formed by the perpendicular line segments, we first need to visualize the problem.
The curve y = 2√x represents a half of a parabola, where the vertex is at (0, 0) and opens upward. The line segments of length h = y are drawn perpendicular to the xy-plane. We are asked to find the area of the surface formed by these line segments from the point (0, 0) to the point (3, 2√3).
To solve this problem, we can approximate the area using a series of rectangles and then take the limit as the width of the rectangles approaches zero (doing this will give us a better approximation and eventually the exact area).
1. Divide the interval [0, 3] into n equal subintervals.
Let's call the width of each subinterval Δx = 3/n.
2. For each subinterval, find the corresponding y-value on the curve y = 2√x.
Let's call this y-value yᵢ, where i is the index of the subinterval.
3. Construct a rectangle on each subinterval with width Δx and height yᵢ.
The area of each rectangle is ΔAᵢ = Δx * yᵢ.
4. Sum up all the areas of the rectangles.
The approximate area is A(n) = ΣΔAᵢ = Σ(Δx * yᵢ) for i = 1 to n.
5. Take the limit as n approaches infinity to find the exact area.
The area of the surface is A = lim(n→∞) A(n).
To evaluate the limit, we can rewrite the sum as a Riemann sum:
A = lim(n→∞) Σ(Δx * yᵢ) = ∫₀³ 2√x dx.
Integrating the function 2√x with respect to x from 0 to 3 will give us the exact area of the surface formed by the perpendicular line segments from (0, 0) to (3, 2√3):
A = ∫₀³ 2√x dx = [4/3 * x^(3/2)] from 0 to 3.
Evaluating this integral, we get:
A = [4/3 * (3)^(3/2)] - [4/3 * (0)^(3/2)] = [4/3 * 3√3] - [4/3 * 0] = 4√3.
Therefore, the area of the surface formed by the perpendicular line segments is 4√3 square units.