Problem and Solution
We are given a curve $S$ with the equation $$ y = x^2 + 2 $$ and a finite region $R$ bounded by the curve, the $y$-axis, and the line $$ y = a $$ where $$ a > 2 $$
The region $R$ is rotated through $360^\circ$ about the $y$-axis to generate a solid with volume $$ 18\pi $$
We are tasked with finding the value of $a$ using algebraic integration.
Solution
Step 1: Set up the volume formula for a solid of revolution
The volume of a solid generated by rotating a region about the $y$-axis is
$$ V = \pi \int_{y_1}^{y_2} x^2 \, dy $$
where:
- $x(y)$ is the radius at height $y$
- $y_1$ and $y_2$ are the limits of integration
The curve is
$$ y = x^2 + 2 $$
so solving for $x$:
$$ x = \sqrt{y - 2} $$
The limits of integration are from $$ y = 2 \quad \text{to} \quad y = a $$
Thus,
$$ V = \pi \int_{2}^{a} \left( \sqrt{y - 2} \right)^2 \, dy = \pi \int_{2}^{a} (y - 2) \, dy $$
Step 2: Compute the integral
Now evaluate
$$ V = \pi \int_{2}^{a} (y - 2) \, dy $$
The antiderivative of $$ y - 2 $$ is
$$ \frac{(y - 2)^2}{2} $$
So,
$$ V = \pi \left[ \frac{(y - 2)^2}{2} \right]_{2}^{a} $$
Substitute the limits:
$$ V = \pi \left( \frac{(a - 2)^2}{2} - 0 \right) $$
$$ V = \frac{\pi}{2} (a - 2)^2 $$
Step 3: Set the volume equal to $18\pi$ and solve for $a$
We are given
$$ V = 18\pi $$
So,
$$ \frac{\pi}{2} (a - 2)^2 = 18\pi $$
Divide both sides by $\pi$:
$$ \frac{1}{2} (a - 2)^2 = 18 $$
Multiply both sides by $2$:
$$ (a - 2)^2 = 36 $$
Take square roots:
$$ a - 2 = 6 \quad \text{or} \quad a - 2 = -6 $$
Thus,
$$ a = 8 \quad \text{or} \quad a = -4 $$
Since $$ a > 2 $$ we take
$$ a = 8 $$
Final Answer
The value of $a$ is
$$ \boxed{8} $$
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