Find the exact area of the surface obtained by rotating about the -axis the
curve: .

Ask by Pritchard Weaver. in China

Dec 24,2024

Question

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

UpStudy AI · Accepted Answer

To find the exact area of the surface obtained by rotating the curve $$ x = \frac{1}{3}\left(y^{2}+2\right)^{\frac{3}{2}} $$ about the $$ x $$-axis for $$ 1 \leq y \leq 2 $$, we'll use the formula for the surface area of a curve described parametrically in terms of $$ y $$:

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

**Step 1: Compute $$ \frac{dx}{dy} $$**

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

Differentiate with respect to $$ y $$:
$$
\frac{dx}{dy} = \frac{1}{3} \cdot \frac{3}{2}(y^2 + 2)^{\frac{1}{2}} \cdot 2y = y(y^2 + 2)^{\frac{1}{2}}
$$

**Step 2: Compute the integrand**

$$
\left(\frac{dx}{dy}\right)^2 + 1 = y^2(y^2 + 2) + 1 = y^4 + 2y^2 + 1 = (y^2 + 1)^2
$$

Taking the square root:
$$
\sqrt{\left(\frac{dx}{dy}\right)^2 + 1} = y^2 + 1
$$

**Step 3: Set up the integral**

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

**Step 4: Evaluate the integral**

$$
\int (y^3 + y) \, dy = \frac{y^4}{4} + \frac{y^2}{2}
$$

Evaluate from 1 to 2:
$$
\left[\frac{2^4}{4} + \frac{2^2}{2}\right] - \left[\frac{1^4}{4} + \frac{1^2}{2}\right] = \left[4 + 2\right] - \left[\frac{1}{4} + \frac{1}{2}\right] = 6 - \frac{3}{4} = \frac{21}{4}
$$

**Step 5: Multiply by $$ 2\pi $$ to get the surface area**

$$
S = 2\pi \cdot \frac{21}{4} = \frac{21}{2} \pi
$$

**Final Answer:**
$$
\boxed{\dfrac{21}{2}\pi}
$$
The exact area of the surface obtained by rotating the curve about the $$ x $$-axis is $$ \frac{21}{2}\pi $$.

Find the exact area of the surface obtained by rotating about the -axis the
curve: .

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Find the exact area of the surface obtained by rotating about the -axis the curve: .