Question
Find the volume generated when the region bounded by \( f(x)=\frac{8}{\sqrt{x}} \) for \( 1 \leq x \leq 8 \) is rotated through \( 2 \pi \) about the \( x \)-axis. A \( \quad \pi \int_{1}^{8} \frac{64}{\sqrt{x}} d x \) B \( \pi \int_{1}^{8} \frac{8}{x} d x \) C \( \pi \int_{1}^{8} \frac{64}{x} d x \) D \( \pi \int_{1}^{8} \frac{8}{\sqrt{x}} d x \)
Ask by Crawford Clarke. in Malaysia
Mar 08,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The correct answer is \( \pi \int_{1}^{8} \frac{64}{x} \, dx \), which corresponds to option C.
Solution
We are given the function
\[
f(x)= \frac{8}{\sqrt{x}},
\]
defined on the interval \(1 \leq x \leq 8\). The volume \( V \) of the solid formed by rotating this region about the \( x \)-axis is given by the formula
\[
V = \pi \int_{a}^{b} \left[f(x)\right]^2 dx.
\]
Substitute \( f(x) \) into the formula:
\[
V = \pi \int_{1}^{8} \left(\frac{8}{\sqrt{x}}\right)^2 dx.
\]
Square the function:
\[
\left(\frac{8}{\sqrt{x}}\right)^2 = \frac{64}{x}.
\]
Thus, the volume becomes:
\[
V = \pi \int_{1}^{8} \frac{64}{x} \, dx.
\]
Comparing this expression with the given options, it matches option C:
\[
\pi \int_{1}^{8} \frac{64}{x} \, dx.
\]
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Simplify this solution Bonus Knowledge
To find the volume generated when the region bounded by \( f(x) = \frac{8}{\sqrt{x}} \) for \( 1 \leq x \leq 8 \) is rotated about the x-axis, we use the disk method. The volume \( V \) is given by the formula: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] In this case, we first square the function \( f(x) \): \[ [f(x)]^2 = \left( \frac{8}{\sqrt{x}} \right)^2 = \frac{64}{x} \] Now, substituting \( f(x)^2 \) into the volume formula with \( a = 1 \) and \( b = 8 \): \[ V = \pi \int_{1}^{8} \frac{64}{x} \, dx \] Thus, the correct expression for the volume is: **C** \( \pi \int_{1}^{8} \frac{64}{x} \, dx \)
