Find the surface area for the solid of revolution obtained by rotating $y=e^{2 x}$ around the $x$-axis over the interval $[-1,2]$. Round your answer to the nearest thousandth.
Answer (2)
Apply the surface area formula: S = 2 π ∫ a b y 1 + ( d x d y ) 2 d x .
Calculate the derivative: d x d y = 2 e 2 x .
Substitute and simplify the integral: S = 2 π ∫ − 1 2 e 2 x 1 + 4 e 4 x d x .
Evaluate the integral and round to the nearest thousandth: 9369.149 .
Explanation
Problem Setup We are asked to find the surface area of the solid of revolution obtained by rotating the curve y = e 2 x around the x -axis over the interval [ − 1 , 2 ] . We need to use the formula for the surface area of a solid of revolution to solve this problem.
Surface Area Formula The formula for the surface area of a solid of revolution about the x -axis is given by: S = 2 π ∫ a b y 1 + ( d x d y ) 2 d x where y = f ( x ) is the curve being rotated, and [ a , b ] is the interval over which the curve is rotated.
Compute the Derivative and Substitute First, we need to find the derivative of y = e 2 x with respect to x :
d x d y = d x d ( e 2 x ) = 2 e 2 x Now, we substitute y = e 2 x and d x d y = 2 e 2 x into the surface area formula: S = 2 π ∫ − 1 2 e 2 x 1 + ( 2 e 2 x ) 2 d x = 2 π ∫ − 1 2 e 2 x 1 + 4 e 4 x d x
Perform u-Substitution To evaluate the integral, we can use the substitution u = 2 e 2 x . Then, d u = 4 e 2 x d x , so e 2 x d x = 4 1 d u . When x = − 1 , u = 2 e − 2 . When x = 2 , u = 2 e 4 . The integral becomes: S = 2 π ∫ 2 e − 2 2 e 4 1 + u 2 4 1 d u = 2 π ∫ 2 e − 2 2 e 4 1 + u 2 d u
Apply the Integral Formula We can use the integral formula: ∫ 1 + u 2 d u = 2 1 [ u 1 + u 2 + sinh − 1 ( u ) ] + C So, the surface area is: S = 2 π [ 2 1 [ u 1 + u 2 + sinh − 1 ( u ) ] ] 2 e − 2 2 e 4 = 4 π [ ( 2 e 4 ) 1 + 4 e 8 + sinh − 1 ( 2 e 4 ) − ( 2 e − 2 ) 1 + 4 e − 4 − sinh − 1 ( 2 e − 2 ) ]
Approximate the Value Now we approximate the value of the expression. We have: S = 4 π [ ( 2 e 4 ) 1 + 4 e 8 + sinh − 1 ( 2 e 4 ) − ( 2 e − 2 ) 1 + 4 e − 4 − sinh − 1 ( 2 e − 2 ) ] ≈ 4 π [ ( 2 × 54.598 ) 1 + 4 × 54.59 8 4 + sinh − 1 ( 2 × 54.598 ) − ( 2 × 0.135 ) 1 + 4 × 0.13 5 4 − sinh − 1 ( 2 × 0.135 ) ] ≈ 4 π [ 109.196 × 1 + 44784416 + sinh − 1 ( 109.196 ) − 0.270 × 1 + 0.0034 − sinh − 1 ( 0.270 ) ] ≈ 4 π [ 109.196 × 6692.863 + 5.386 − 0.270 × 1.0017 − 0.266 ] \approx \frac{\pi}{4} \left[ 730840.15 + 5.386 - 0.270 - 0.266 \right] \approx \frac{\pi}{4} \left[ 730844.999 \right] \approx 9369.149
Final Answer Therefore, the surface area of the solid of revolution is approximately 9369.149 .
Examples
Surface area calculations are crucial in various engineering applications, such as designing heat exchangers. For instance, when optimizing the cooling efficiency of a heat sink, engineers need to accurately calculate the surface area to maximize heat dissipation. By rotating a specific curve around an axis, they can create complex shapes with large surface areas, enhancing the cooling performance. This ensures that electronic components operate within safe temperature limits, improving the reliability and lifespan of the devices.
The surface area of the solid formed by rotating the curve y = e 2 x around the x -axis over the interval [ − 1 , 2 ] is approximately 9369.149 , calculated using the surface area formula for solids of revolution. This involves calculating the derivative, setting up the integral, and using numerical methods for evaluation.
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