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Let R be the region bounded by the following curves. Use the disk method to find the volume of the solid generated when R is revolved about the x-axis.

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

also separately revolve around $y=-1, y=2$.

Im having trouble seeing what the radius would be in each case and what shape the revolutions make so I can find an area function.

Thanks

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    The radius of revolution is always the top curve minus the bottom curve when revolving about the x-axis. – 2017-02-03
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    First draw the curves, then obtain the planar region R. Now try to think to rotate R about x-axis. Here is one animation that will help you to think.http://web.calstatela.edu/curvebank/volrev/rev1.gif – 2017-02-03

1 Answers 1

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The key is to sketch the diagram.

solids of revolution

Using the disk/annulus method we see that a plane section of the solid of revolution taken perpendicular to the axis of revolution intersects the solid in an annulus with an outer radius $R$ and inner radius $r$.

Thus we use the formula for the area of an annulus to calculate the volumn.

\begin{equation} V=\int_0^{\frac{\pi}{4}}\pi\left(R^2-r^2\right)\,dx \end{equation}

For the solid on the left we see from the diagram that $R=2$ and $r=\sec(x)$ therefore

\begin{equation} V=\pi \int_0^{\frac{\pi}{4}}4-\sec^2(x)\,dx \end{equation}

For the solid on the right we see that $R=1+\sec(x)$ and $r=1$ therefore

\begin{equation} V=\pi \int_0^{\frac{\pi}{4}}(1-\sec(x))^2-1\,dx \end{equation}

Both are elementary integrals involving $\int \sec(x)\,dx$ and $\int \sec^2(x)\,dx$

I have provided a desmos annimation at https://www.desmos.com/calculator/gj0cf5rgxt where you can move the axis of revolution between values of $y=k$ for $k\in[-1,2]$ and where you can move the annular cross-section between values of $x=c$ for $c\in\left[0,\frac{\pi}{4}\right]$ by moving the sliders for $k$ and $c$.