The region between the curve $$ y=\sec x, 0 \leq x \leq \frac{\pi}{4} $$ and the $$ x $$-axis is revolved about the $$ x $$-axis to generate a solid. Find the exact value of its volume. $$ \text { [?] } $$

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We use the disk method. The radius of a representative disk is given by the function value $$ y = \sec x $$. The volume $$ V $$ of the solid obtained by revolving the region about the $$ x $$-axis is

$$
V = \pi \int_{0}^{\frac{\pi}{4}} \left(\sec x\right)^2 \, dx.
$$

We know that

$$
\int \sec^2 x \, dx = \tan x.
$$

Thus, the volume is

$$
V = \pi \left[ \tan x \right]_0^{\frac{\pi}{4}} = \pi \left(\tan \frac{\pi}{4} - \tan 0\right).
$$

Since

$$
\tan \frac{\pi}{4} = 1 \quad \text{and} \quad \tan 0 = 0,
$$

we have

$$
V = \pi (1 - 0) = \pi.
$$

The exact value of the volume is $$\pi$$.
The volume of the solid is $$\pi$$.

The region between the curve

and the -axis is revolved about
the -axis to generate a solid.
Find the exact value of its volume.

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The region between the curve and the -axis is revolved about the -axis to generate a solid. Find the exact value of its volume.