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Chapter 6: Problem 13

Let \(R\) be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about the \(y\) -axis. $$y=\cos x^{2}, y=0, \text { for } 0 \leq x \leq \sqrt{\pi / 2}$$

Short Answer

Expert verified
Answer: The volume of the solid is 蟺 cubic units.

Step by step solution

01

Identifying the radius and height of the cylindrical shell

The radius of a cylindrical shell is the distance from the y-axis to a point on the curve y = cos(x^2). As x ranges from 0 to \(\sqrt{\pi / 2}\), the radius is given by x, which can also be represented as the integral variable, say u. The height of the cylindrical shell is the difference between the two curve values. In this case, it's the difference between \(\cos(x^2)\) and 0. So, the height is given by \(\cos(u^2)\), where u is the same as x.
02

Writing the volume of a cylindrical shell formula

The volume of a cylindrical shell is given by: \(V = 2 \pi \int_{a}^{b} r(u)h(u) du\) In our case, a = 0 and b = \(\sqrt{\pi / 2}\), r(u) = u, and h(u) = \(\cos(u^2)\). Therefore, the volume of the solid can be expressed as: \(V = 2 \pi \int_{0}^{\sqrt{\pi/2}} u\cos(u^2) du\)
03

Computing the integral

To evaluate the integral, we will use substitution. Let \(v = u^2\). Then, we have \(\frac{dv}{du} = 2u\), so \(du = \frac{dv}{2u}\). Also, when u = 0, v = 0, and when u = \(\sqrt{\pi/2}\), v = \(\pi/2\). Now, we can replace the variables and the limits of integration with our substitution. \(V = 2 \pi \int_{0}^{\pi/2} \frac{1}{2} \cos(v) dv\). Simplify the expression for the volume: \(V = \pi \int_{0}^{\pi/2} \cos(v) dv\). Now, integrate the expression: \(V = \pi \left[ \sin(v) \right]_0^{\pi/2}\).
04

Evaluating the final expression

Evaluate the definite integral to obtain the volume of the solid: \(V = \pi(\sin(\pi / 2) - \sin(0)) = \pi(1-0) = \pi\). Thus, the volume of the solid generated when the region R is revolved about the y-axis is \(\pi\) cubic units.

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