Find The Area Of The Surface Generated When The Given Curve Is Revolved About The Given Ax Y=4sqrt(x),

Find The Area Of The Surface Generated When The Given Curve Is Revolved About The Given Axis: Y = 4√x

Understanding the surface area generated by the revolution of a curve around an axis is a fundamental concept in calculus. It combines the principles of integration with geometric intuition to determine the surface area of a three-dimensional shape formed by rotating a two-dimensional curve. In this article, we focus on the specific problem: finding the surface area generated when the curve y = 4√x is revolved about a given axis, typically the x-axis or y-axis.

This problem is not only mathematically interesting but also has practical applications in engineering, physics, and manufacturing, where surface area calculations are essential for designing objects, calculating material usage, and analyzing physical phenomena.

Understanding the Problem: The Curve y = 4√x

Before delving into the mathematics, it’s important to understand the characteristics of the given curve:

• The equation y = 4√x describes a square root function scaled vertically by a factor of 4.


• The domain of the function is x ≥ 0, as the square root function is defined for non-negative x.


• The range of the function is y ≥ 0, since the square root yields non-negative results.


• The curve starts at the origin (0, 0) and increases gradually as x increases.

Visualizing this curve helps in understanding how the surface area will be generated when it is revolved about an axis.

Revolution About the X-Axis and Y-Axis: Key Concepts

The surface area generated by revolving a curve around an axis can vary significantly depending on the axis chosen. The two common axes of revolution are:

1. Revolution About the X-Axis

• The curve y = 4√x is rotated around the x-axis.
• The resulting surface resembles a "bowl" or "shell" extending along the x-axis.
• The surface area depends on the length of the curve and the distance from each point to the axis.

2. Revolution About the Y-Axis

• The curve is revolved around the y-axis.
• The surface resembles a "bottle" or "vase" shape.
• The calculation involves different limits and variables.

For this article, we’ll primarily focus on the revolution about the x-axis, which is the most common scenario in calculus problems involving y = 4√x.

Mathematical Foundation: Surface Area Formula

The general formula for the surface area (SA) generated when a curve y = f(x) is revolved about the x-axis, over the interval [a, b], is:

Where:

• y = f(x) is the curve being revolved.


• dy/dx is the derivative of y with respect to x.


• The limits a and b define the interval over which the surface area is computed.

Similarly, if the rotation is about the y-axis, an analogous formula applies, involving integration with respect to y, but for the current focus, the x-axis is our primary axis.

Step-by-Step Solution: Calculating Surface Area for y = 4√x

To find the surface area of the surface generated when the curve y = 4√x is revolved about the x-axis, follow these steps:

Step 1: Define the interval of interest

• Suppose we are interested in the surface area between x = a and x = b.
• For example, from x = 0 to x = c, where c > 0.

Step 2: Find the derivative dy/dx

Given y = 4√x = 4x^(1/2):

dy/dx = 4 (1/2) x^(-1/2) = 2 x^(-1/2) = 2 / √x

Step 3: Calculate (dy/dx)²

(dy/dx)² = (2 / √x)² = 4 / x

Step 4: Set up the surface area integral

Using the formula:

SA = 2π ∫ₐᵇ y √(1 + (dy/dx)²) dx

Substitute y = 4√x and (dy/dx)² = 4 / x:

SA = 2π ∫ₐᵇ 4√x √(1 + 4 / x) dx

Simplify inside the square root:

1 + 4 / x = (x + 4) / x

Thus,

√(1 + 4 / x) = √((x + 4)/ x) = √(x + 4) / √x

Therefore, the integrand becomes:

4√x (√(x + 4) / √x) = 4 √(x + 4)

Step 5: Rewrite the integral

SA = 2π ∫ₐᵇ 4 √(x + 4) dx = 8π ∫ₐᵇ √(x + 4) dx

Now, the problem reduces to evaluating:

SA = 8π ∫ₐᵇ √(x + 4) dx

Step 6: Integrate √(x + 4) dx

Let u = x + 4, so that du = dx.

When x = a, u = a + 4.

When x = b, u = b + 4.

The integral becomes:

∫ √u du = ∫ u^{1/2} du

The integral of u^{1/2} is:

(2/3) u^{3/2}

Thus,

SA = 8π (2/3) [u^{3/2}] evaluated from u = a + 4 to u = b + 4

SA = (16π/3) [(b + 4)^{3/2} - (a + 4)^{3/2}]

Final Expression for Surface Area

The surface area generated when the curve y = 4√x is revolved about the x-axis over the interval [a, b] is:

This formula allows you to compute the surface area for any chosen interval.

Specific Example: From x = 0 to x = 4

Let’s compute the surface area when x ranges from 0 to 4:

• a = 0, b = 4

Plug into the formula:

SA = (16π/3) [(4 + 4)^{3/2} - (0 + 4)^{3/2}]

Calculate each term:

(8)^{3/2} = (8)^{1} (8)^{1/2} = 8 √8 = 8 2√2 = 16√2

(4)^{3/2} = 4 √4 = 4 2 = 8

Therefore,

SA = (16π/3) (16√2 - 8) = (16π/3) 8 (2√2 - 1)

Simplify:

SA = (16π/3) 8 (2√2 - 1) = (16 8 π/3) (2√2 - 1) = (128π/3) (2√2 - 1)

Hence, the surface area is:

SA = (128π/3) (2√2 - 1)

This numerical value can be approximated as:

• 2√2 ≈ 2 1.4142 ≈ 2.8284


• 2.8284 - 1 = 1.8284


• SA ≈ (128π/3) 1.8284 ≈ (128 3.1416 / 3) 1.8284 ≈ (128 1.0472) 1.8284 ≈ 134.07 1.8284 ≈ 245.31 square units

This comprehensive calculation illustrates how the surface area can be precisely determined.

Extensions and Variations

While the example above considers revolution about the x-axis, similar methods apply if the curve is revolved around the y-axis, or if the interval changes.

Revolution About the Y-Axis

• The formula involves integrating with respect to y.
• Express x in terms of y: x = (y/4)^2
• Derive dx/dy, and set up the integral accordingly.

Other Intervals and Curves

• The approach remains similar: find the derivative, simplify the integrand, and evaluate the integral over the specified limits.