Evaluate whether the following solution to the problem below is accurate:
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{Problem}
Find the surface area of the solid generated by revolving the parametric curve
[
x = t^2, \quad y = \frac{t^3}{3}
]
about the (x)-axis. From t = 0 to t = T (Hint - The integral obtained during the steps is quite complex and requires trigonometric substitution. It can be solved using integration by parts and trigonometric identities. Let t = 2 * tan(u), then dt = 2 * sec²(u) du. Substituting, we get: ...)
Solution 1
To find the surface area (S) of the solid generated by revolving the parametric curve (x = t^2), (y = \frac{t^3}{3}) about the (x)-axis from (t = 0) to (t = T), we will use the formula for the surface area of a parametric curve rotated about the (x)-axis:
[ S = \int{t=0}^{t=T} 2\pi y(t) \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2} , dt ]
Step 1: Compute the derivatives
First, compute the derivatives of (x(t)) and (y(t)):
[ \dfrac{dx}{dt} = 2t, \quad \dfrac{dy}{dt} = t^2 ]
Step 2: Compute the square root term
Compute the expression under the square root:
[ \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2} = \sqrt{(2t)^2 + (t^2)^2} = \sqrt{4t^2 + t^4} = t\sqrt{4 + t^2} ]
Step 3: Set up the integral
Substitute (y(t)) and the expression from Step 2 into the surface area formula:
[ S = \int{t=0}^{t=T} 2\pi \left(\frac{t^3}{3}\right) \cdot t\sqrt{4 + t^2} , dt = \dfrac{2\pi}{3} \int{t=0}^{t=T} t^4 \sqrt{4 + t^2} , dt ]
Step 4: Use trigonometric substitution
Let (t = 2\tan(u)), so (dt = 2\sec^2(u) du), and (\sqrt{4 + t^2} = \sqrt{4 + 4\tan^2(u)} = 2\sec(u)).
Change the limits of integration: when (t = 0), (u = 0); when (t = T), (u = \arctan\left(\dfrac{T}{2}\right)).
Substitute into the integral:
[ S = \dfrac{2\pi}{3} \int{u=0}^{u=\arctan(T/2)} [16\tan^4(u)] [2\sec(u)] [2\sec^2(u) du] = \dfrac{128\pi}{3} \int_{u=0}^{u=\arctan(T/2)} \tan^4(u) \sec^3(u) du ]
Step 5: Simplify the integrand
Expand (\tan^4(u)):
[ \tan^4(u) = (\sec^2(u) - 1)^2 = \sec^4(u) - 2\sec^2(u) + 1 ]
So the integrand becomes:
[ \int (\sec^4(u) - 2\sec^2(u) + 1) \sec^3(u) du = \int \left( \sec^7(u) - 2\sec^5(u) + \sec^3(u) \right) du ]
Step 6: Compute the integrals
We apply reduction formulas for each term:
Integrate (\int \sec^3(u) du):
[ \int \sec^3(u) du = \dfrac{\sec(u)\tan(u)}{2} + \dfrac{1}{2} \ln|\sec(u) + \tan(u)| ]
Integrate (\int \sec^5(u) du):
[ \int \sec^5(u) du = \dfrac{\sec^3(u)\tan(u)}{4} + \dfrac{3}{4} \int \sec^3(u) du ]
Integrate (\int \sec^7(u) du):
[ \int \sec^7(u) du = \dfrac{\sec^5(u)\tan(u)}{6} + \dfrac{5}{6} \int \sec^5(u) du ]
Combining these results and simplifying, we find:
[ \int \left( \sec^7(u) - 2\sec^5(u) + \sec^3(u) \right) du = \dfrac{\sec^5(u)\tan(u)}{6} - \dfrac{7}{24} \sec^3(u) \tan(u) + \dfrac{1}{16} \sec(u)\tan(u) + \dfrac{1}{16} \ln|\sec(u) + \tan(u)| ]
Step 7: Evaluate the definite integral
Convert back to (t) using (u = \arctan\left(\dfrac{t}{2}\right)). We have:
[ \sec(u) = \frac{\sqrt{4 + t^2}}{2}, \quad \tan(u) = \frac{t}{2} ]
So:
[ S = \dfrac{128\pi}{3} \left[ \dfrac{1}{384} \sqrt{4 + T^2}, T (T^4 + T^2 - 6) + \dfrac{1}{16} \ln \left( \dfrac{\sqrt{4 + T^2} + T}{2} \right) \right] ]
Simplify the constants and the expression:
[ S = \dfrac{\pi}{9} \sqrt{4 + T^2}, T (T^4 + T^2 - 6) + \dfrac{8\pi}{3} \ln \left( \dfrac{\sqrt{4 + T^2} + T}{2} \right) ]
Answer:
An explicit formula: Surface area S = (π⁄9)·T·√(4＋T²)·(T⁴ ＋ T² － 6) ＋ (8π⁄3)·ln[(√(4＋T²)＋T) ⁄ 2]
{The Actual Correct Answer}
S=(2π/3) * (T⁷ + 5T⁵ - 2T³ - 24T + 24√(T² + 4)asinh(T/2)) / (6√(T² + 4))