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The region between the curve

y=sec x,0 <= x <= (pi)/(4)
and the
x-axis is revolved about the
x-axis to generate a solid.
Find the exact value of its volume.

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

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Find derivatives of inverse trigonometric functions

Full solution

Q. The region between the curve

\newline

y=\sec x, 0 \leq x \leq \frac{\pi}{4}

\newline

and the

x

-axis is revolved about the

x

-axis to generate a solid.

\newline

Find the exact value of its volume.

Disk Method Integration: To find the volume of the solid, we use the disk method, which involves integrating the area of circular disks along the x-axis from $x=0$ to $x=\frac{\pi}{4}$ .
Area of Disk Calculation: The area of a disk at a particular $x$ is given by $A(x) = \pi[\sec(x)]^2$ , since the radius of the disk is the $y$ -value of the function, which is $\sec(x)$ .
Volume Integral Calculation: The volume $V$ is the integral of $A(x)$ from $x=0$ to $x=\frac{\pi}{4}$ . So, $V = \int_{0}^{\frac{\pi}{4}} \pi[\sec(x)]^2 \, dx$ .
Integral Evaluation: Now we calculate the integral: $V = \pi\int_{0}^{\frac{\pi}{4}} \sec^2(x) \, dx$ .
Final Volume Calculation: The antiderivative of $\sec^2(x)$ is $\tan(x)$ , so we evaluate $\tan(x)$ from $0$ to $(\pi)/(4)$ .
Final Volume Calculation: The antiderivative of $\sec^2(x)$ is $\tan(x)$ , so we evaluate $\tan(x)$ from $0$ to $\frac{\pi}{4}$ .Evaluating the antiderivative at the bounds gives us $V = \pi[\tan(\frac{\pi}{4}) - \tan(0)]$ .
Final Volume Calculation: The antiderivative of $\sec^2(x)$ is $\tan(x)$ , so we evaluate $\tan(x)$ from $0$ to $(\pi)/(4)$ . Evaluating the antiderivative at the bounds gives us $V = \pi[\tan((\pi)/(4)) - \tan(0)]$ . Since $\tan((\pi)/(4)) = 1$ and $\tan(0) = 0$ , we have $V = \pi[1 - 0]$ .
Final Volume Calculation: The antiderivative of $\sec^2(x)$ is $\tan(x)$ , so we evaluate $\tan(x)$ from $0$ to $(\pi)/(4)$ . Evaluating the antiderivative at the bounds gives us $V = \pi[\tan((\pi)/(4)) - \tan(0)]$ . Since $\tan((\pi)/(4)) = 1$ and $\tan(0) = 0$ , we have $V = \pi[1 - 0]$ . Therefore, the volume $V = \pi$ cubic units.

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