Question

In Exercises 39-42, find the volume of the solid analytically. The solid lies between planes perpendicular to the \(x\) -axis at \(x=-\pi / 3\) and \(x=\pi / 3 .\) The cross sections perpendicular to the \(x\) -axis are (a) circular disks with diameters running from the curve \(y=\tan x\) to the curve \(y=\sec x\). (b) squares whose bases run from the curve \(y=\tan x\) to the curve \(y=\sec x\).

Short answer

The answers are the values of the integrals calculated in steps 2 and 4.

Step-by-step solution

Set up the Integrand for Part (a)

We are dealing with discs, so the formula for the volume of the solid is given by \(\int_{a}^{b} A(x) dx\), where \(A(x)\) is the area of the cross section at \(x\). The value \(A(x)\) is given by the formula for the area of a circle, \(A(x) = \pi r^2\), where \(r\) is the radius of the disc. In our case, the radius \(r\) is the difference of the equations \(y=\sec x\) and \(y=\tan x\). So, \(A(x) = \pi (\sec x - \tan x)^2\).

Evaluate the Integral for Part (a)

We evaluate the definite integral \(\int_{-\pi / 3}^{\pi / 3} \pi (\sec x - \tan x)^2 dx\) to find the volume of the solid.

Set up the Integrand for Part (b)

Now we are dealing with squares. The formula for the volume of the solid remains the same, but this time, \(A(x)\) is \((\sec x - \tan x)^2\), which is the area of a square with side length equal to the difference between the curves \(y=\sec x\) and \(y=\tan x\).

Evaluate the Integral for Part (b)

We evaluate the definite integral \(\int_{-\pi / 3}^{\pi / 3} (\sec x - \tan x)^2 dx\) to find the volume of the solid.