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.
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 S 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.
We can calculate the area of this revolution in various ways such as:
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
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
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 = ex 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.
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