Surface Area of a Solid of Revolution

Surface Area of a Solid of Revolution: The area of the surface of a solid of revolution is a fundamental concept in calculus and engineering. It involves rotating a curve around an axis to form a three-dimensional solid and then determining the surface area of this solid.

This concept has significant applications in various fields, including physics, engineering, and manufacturing. In this article, we will discuss the surface area of a solid of revolution, its formula, examples, applications in engineering, etc.

Surface Area when Rotating Around the X-Axis

Consider a function y = f(x) that is continuous and differentiable on the interval [a,b]. The surface area S of the solid of revolution obtained by rotating this curve around the x-axis is given by:

S = 2\pi \int_a^b f(x) \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx

Surface Area when Rotating Around the Y-Axis

Consider a function x=g(y) that is continuous and differentiable on the interval [c,d]. The surface area SSS of the solid of revolution obtained by rotating this curve around the y-axis is given by:

S = 2\pi \int_c^d g(y) \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy

Consider a plane y=f(x) in the x-y plane between ordinates x=a and x=b. A solid of revolution is generated if a certain portion of this curve is revolved about an axis.

Read More: Volume of Revolution - Definition, Formula, and Examples

We can calculate the area of this revolution in various ways such as:

Cartesian Form:

Area of solid formed by revolving the arc of the curve about the x-axis is-S= \int_{x=a}^{x=b} 2\pi y\sqrt{1+(\frac{dy}{dx})^2}dx

Area of revolution by revolving the curve about y-axis is-S= \int_{y=c}^{y=d} 2\pi x \sqrt{1+(\frac{dx}{dy})^2}dy

Parametric Form: x=x(t), y=y(t)

About x-axis:S=\int_{t=t_{1}}^{t=t_{2}} 2\pi y(t) \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt

About y-axis:S=\int_{t=t_{1}}^{t=t_{2}} 2\pi x(t) \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt

Polar Form: r=f(θ)

About the x-axis: initial line \theta = \frac{\pi}{2} S= \int_{\theta=\theta_1}^{\theta _2}2\pi y\frac{ds}{d\theta}d\theta=\int_{\theta=\theta_1}^{\theta _2}2\pi (r sin\theta) \sqrt{r^2+(\frac {dr}{d\theta})^2}d\theta Here replace r by f(θ)

About the y-axis:S= \int_{\theta=\theta_1}^{\theta _2}2\pi x\frac{ds}{d\theta}d\theta=\int_{\theta=\theta_1}^{\theta _2}2\pi (r cos\theta) \sqrt{r^2+(\frac {dr}{d\theta})^2}d\theta Here replace r by f(θ)

About any axis or line L:

S= \int 2\pi (PM) ds where PM is the perpendicular distance of a point P of the curve to the given axis.

Limits for x: x = a to x = bS=\int_{x=a}^{x=b} 2\pi (PM)\sqrt{1+(\frac{dy}{dx})^2}dx Here PM is in terms of x.

Limits for y: y = c to y = dS= \int_{y=c}^{y=d} 2\pi (PM)\sqrt{1+(\frac{dx}{dy})^2}dy Here PM is in terms of y.

Solved examples :

Example 1:

Find the surface area when y = 2x is revolved about the x-axis from x = 0 to x = 3.

Solution:

f(x) = 2x

f'(x) = 2

S = 2π ∫[0 to 3] 2x √(1 + 2²) dx

= 2π ∫[0 to 3] 2x √5 dx

= 2π√5 ∫[0 to 3] 2x dx

= 2π√5 [x²][0 to 3]

= 2π√5 (9 - 0) = 18π√5 ≈ 126.65 square units

Example 2:

Find the surface area when y = √x is revolved about the x-axis from x = 0 to x = 4.

Solution:

f(x) = √x

f'(x) = 1/(2√x)

S = 2π ∫[0 to 4] √x √[1 + 1/(4x)] dx

= 2π ∫[0 to 4] √x √[(4x + 1)/(4x)] dx

= 2π ∫[0 to 4] √[(x(4x + 1))/4] dx

= π ∫[0 to 4] √(4x² + x) dx

= π [(2x²/3 + x/2) √(4x² + x) + (1/8) ln(√(4x² + x) + 2x)][0 to 4]

≈ 70.21 square units

Example 3:

Find the surface area when y = x² is revolved about the x-axis from x = 0 to x = 2.

Solution:

f(x) = x²

f'(x) = 2x

S = 2π ∫[0 to 2] x² √(1 + 4x²) dx

= 2π [x³√(1 + 4x²)/3 + (1/24)ln(2x + √(1 + 4x²))][0 to 2]

≈ 45.35 square units

Example 4:

Find the surface area when y = sin(x) is revolved about the x-axis from x = 0 to x = π.

Solution:

f(x) = sin(x)

f'(x) = cos(x)

S = 2π ∫[0 to π] sin(x) √(1 + cos²(x)) dx

= 2π ∫[0 to π] sin(x) √(2 - sin²(x)) dx

= 2π [√2 - 2][0 to π] ≈ 26.32 square units

Example 5:

Find the surface area when y = e^x is revolved about the x-axis from x = 0 to x = 1.

Solution:

f(x) = e^x

f'(x) = e^x

S = 2π ∫[0 to 1] e^x √(1 + e^(2x)) dx

= 2π [√(1 + e^(2x))/2][0 to 1]

= π [√(1 + e²) - √2] ≈ 23.82 square units

Example 6:

Find the surface area when y = 3 - x² is revolved about the x-axis from x = -1 to x = 1.

Solution:

f(x) = 3 - x²

f'(x) = -2x

S = 2π ∫[-1 to 1] (3 - x²) √(1 + 4x²) dx

= 2π [3x√(1 + 4x²)/2 - x³√(1 + 4x²)/6 + (3/8)arcsinh(2x)][-1 to 1]

≈ 24.13 square units

Example 7:

Find the surface area when y = x³ is revolved about the x-axis from x = 0 to x = 2.

Solution:

f(x) = x³

f'(x) = 3x²

S = 2π ∫[0 to 2] x³ √(1 + 9x⁴) dx

This integral doesn't have an elementary antiderivative. We can evaluate it numerically:

S ≈ 67.02 square units

Example 8:

Find the surface area when y = ln(x) is revolved about the x-axis from x = 1 to x = e.

Solution:

f(x) = ln(x)

f'(x) = 1/x

S = 2π ∫[1 to e] ln(x) √(1 + 1/x²) dx

= 2π [x ln(x) √(1 + 1/x²) - ∫ √(1 + 1/x²) dx][1 to e]

= 2π [x ln(x) √(1 + 1/x²) - x√(1 + 1/x²) + arcsinh(1/x)][1 to e]

≈ 30.68 square units

Example 9:

Find the surface area when y = 1/x is revolved about the x-axis from x = 1 to x = 2.

Solution:

f(x) = 1/x

f'(x) = -1/x²

S = 2π ∫[1 to 2] (1/x) √(1 + 1/x⁴) dx

= 2π [√(x² + 1)/x][1 to 2]

= 2π (√5/2 - √2) ≈ 5.13 square units

Example 10:

Find the surface area when y = cos(x) is revolved about the x-axis from x = 0 to x = π/2.

Solution:

f(x) = cos(x)

f'(x) = -sin(x)

S = 2π ∫[0 to π/2] cos(x) √(1 + sin²(x)) dx

= 2π ∫[0 to π/2] cos(x) √(2 - cos²(x)) dx

= 2π [√2 - 1] ≈ 7.19 square units

Applications in Engineering

1. Mechanical Engineering: In mechanical engineering, calculating the surface area of solids of revolution is crucial for designing and manufacturing components such as pipes, tanks, and rotating machinery.

2. Aerospace Engineering: In aerospace engineering, determining the surface area of aerodynamic surfaces helps in analyzing drag and optimizing the design of aircraft and spacecraft.

3. Civil Engineering: In civil engineering, surface area calculations are essential for designing structures such as columns, domes, and arches to ensure stability and structural integrity.

4. Manufacturing: In manufacturing, understanding the surface area of components aids in material estimation, cost analysis, and quality control during production processes.

Conclusion - Surface Area of a Solid of Revolution

The surface area of a solid of revolution is a vital concept in calculus and engineering, providing essential tools for designing and analyzing various structures and components. Understanding the methods to calculate these areas enables engineers to optimize designs and ensure the functionality and efficiency of their creations.

• Volume of Solid of Revolution
• Surface Area of a Solid of Revolution