Find the area of the surface obtained by rotating the curve $x = \frac{1}{3}(y^2 + 2)^{3/2}$, $0 \leq y \leq 3$, about the $x$-axis.

Neetesh Kumar | November 15, 2024

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This is the solution to DHW Calculus
Assignment: 7 Question Number 10
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Step-by-step solution:

Step 1: Surface Area Formula

For rotation about x-axis when curve is in form $x = f(y)$:

$S = 2\pi\int_{a}^{b} y\sqrt{1 + (\frac{dx}{dy})^2}dy$

Step 2: Find $\frac{dx}{dy}$

$x = \frac{1}{3}(y^2 + 2)^{3/2}$

$\frac{dx}{dy} = \frac{1}{3} \cdot \frac{3}{2}(y^2 + 2)^{1/2} \cdot 2y$

$= y\sqrt{y^2 + 2}$

Step 3: Calculate $1 + (\frac{dx}{dy})^2$

$1 + (\frac{dx}{dy})^2 = 1 + y^2(y^2 + 2)$

$= 1 + y^4 + 2y^2$

$= y^4 + 2y^2 + 1$

$= (y^2 + 1)^2$

Step 4: Set up and solve the integral

$S = 2\pi\int_{0}^{3} y\sqrt{(y^2 + 1)^2}dy$

$= 2\pi\int_{0}^{3} y(y^2 + 1)dy$

$= 2\pi\int_{0}^{3} (y^3 + y)dy$

$= 2\pi[\frac{y^4}{4} + \frac{y^2}{2}]_{0}^{3}$

$= 2\pi(\frac{81}{4} + \frac{9}{2})$

$= 2\pi(\frac{81}{4} + \frac{18}{4})$

$= 2\pi \cdot \frac{99}{4}$

Final Answer:

The surface area is: $\boxed{\frac{99\pi}{2}}$

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