Question

Use Wallis's Formula to find the volume of the solid bounded by the graphs of the equations. $$ z=\sin ^{2} x, z=0,0 \leq x \leq \pi, 0 \leq y \leq 5 $$

Short answer

The volume is calculated by solving the integral. The exact value is subject to the correct calculation of the integral.

Step-by-step solution

Understanding the Context

First, identify the bounds and functions. We are given four conditions: \(z=\sin^{2}x\), \(z=0\), \(0 \leq x \leq \pi\), \(0 \leq y \leq 5\). These conditions indicate the volume is over the region defined by \(x\) from 0 to \(\pi\) and \(y\) from 0 to 5, bounded above by \(z=\sin^{2}x\) and below by \(z=0\).

Setup the integral using Wallis's Formula

As the solid can be interpreted as rotating the area between \(z=\sin^{2}x\) and \(z=0\) (from \(x=0\) to \(x=\pi\)) around \(x\)-axis, we then set up the double integral with respect to \(x\) and \(y\) for calculating the volume. The general Wallis's formula for volume in this case is \[V=\int_{a}^{b}\int_{c}^{d}\sin^{2}x dy dx\] where [a,b] is the interval of \(x\) and [c,d] is the interval for \(y\)

Compute the Integral

The inner integral does not involve \(y\), therefore we have \[\int_{0}^{5} dy = 5 \nonumber\] Hence, substituting this into the original integral we get \[V=5\int_{0}^{\pi}\sin^{2}x dx \nonumber\] Next, we use the power reduction identity \(\sin^{2}x = \frac{1-\cos(2x)}{2}\) This simplifies the integral to \[V = 5\int_{0}^{\pi} \frac{1-\cos(2x)}{2} dx\] Solve for \(V\) by integrating term by term.

Final Calculation

Calculate the answer by performing the integral calculation. The result of the integration would give the volume.

Key concepts

Double Integral

Double integrals extend the concept of a single integral, which we would use to find the area under a curve, to finding the volume under a surface in two dimensions. In the context of the given exercise, a double integral is what allows us to calculate the volume of a solid by integrating over two variables: typically, these are along the x-axis and y-axis.

Double integrals are written in the form \(\int_{a}^{b}\int_{c}^{d} f(x,y) dy dx\), where \(f(x,y)\) represents a function over a region in the xy-plane, and the limits \(a, b, c, d\) define the bounds of the region. We integrate \(f(x,y)\) with respect to \(y\) first (the inner integral), and then with respect to \(x\) (the outer integral).

In our specific exercise, the function to be integrated is \(z = \sin^2 x\), which doesn't depend on \(y\). Consequently, the integral with respect to \(y\) becomes a simple multiplication by the range of \(y\), stretching from 0 to 5.

Solid of Revolution

A solid of revolution is a three-dimensional object obtained by revolving a two-dimensional plane area around an axis. This process is similar to turning clay on a potter's wheel to create a vase or a bowl.

In our exercise, we effectively rotate the area between the curve \(z = \sin^2 x\) and \(z = 0\), from \(x = 0\) to \(x = \pi\), around the x-axis. The resulting solid is symmetrical and its volume can be analyzed through an integral, with Wallis's formula serving as a guide.

Visualization

Imagine slicing the solid perpendicular to the axis of revolution into infinitesimally thin disks. The radius of each disk in this scenario is determined by the value of \(z = \sin^2 x\) at that specific \(x\), and the volume of each disk is proportional to its radius squared. Integrating across these slices, we obtain the volume of the entire solid.

Power Reduction Identity

Integrating functions like \(\sin^2 x\) directly can be complex. That's where the power reduction identity becomes beneficial. The power reduction identity is a trigonometric identity that simplifies the integral of a power of a sine or cosine function.

The identity for \(\sin^2 x\) is \(\sin^2x = \frac{1-\cos(2x)}{2}\). By applying this identity, we reduce the power from \(\sin^2 x\) to a linear combination of simpler functions, making the integral more manageable.

In the given exercise, substituting \(\sin^2 x\) with the power reduction identity transforms the task into integrating a constant and the cosine function, both of which have straightforward antiderivatives. Hence, we get to the crux of solving the integral and, consequently, finding the desired volume of the solid of revolution.