Question

Let \(R\) be the region bounded by \(y=\sin x\) and \(y=\cos x\) between \(x=-\pi / 4\) and \(x=3 \pi / 4\). Find the volume of the solid obtained when \(R\) is revolved about \(x=-\pi / 4 .\) Hint: Use cylindrical shells to write a single integral, make the substitution \(u=x-\pi / 4\), and apply symmetry properties.

Short answer

Volume is \(2\pi^2\sqrt{2}\).

Step-by-step solution

Set Up the Problem

We need to find the volume of the solid formed by revolving the region between the curves \( y = \sin x \) and \( y = \cos x \) around the line \( x = -\pi/4 \). We'll use the cylindrical shells method, which involves evaluating an integral of the form \( V = \int 2\pi \cdot (\text{radius}) \cdot (\text{height}) \, dx \). Here the radius is the distance from \( x \) to \( -\pi/4 \) and the height is the vertical distance between the two curves.

Define Variables and Integral

The cylindrical shell radius is \( x - (-\pi/4) = x + \pi/4 \). The height is \( \cos x - \sin x \). So, the volume integral is set up as: \[ V = \int_{-\pi/4}^{3\pi/4} 2\pi(x + \pi/4)(\cos x - \sin x) \, dx \]

Make the Substitution

The problem suggests using the substitution \( u = x - \pi/4 \). Therefore, \( du = dx \) and when \( x = -\pi/4 \), \( u = 0 \); when \( x = 3\pi/4 \), \( u = \pi \). The integral boundaries change accordingly, and the integral re-frames to: \[ V = \int_{0}^{\pi} 2\pi(u + 0)(\cos(u + \pi/4) - \sin(u + \pi/4)) \, du \]

Use Trigonometric Identities

Use trigonometric angle addition formulas: \( \cos(u + \pi/4) = \frac{1}{\sqrt{2}}(\cos u - \sin u) \) \( \sin(u + \pi/4) = \frac{1}{\sqrt{2}}(\cos u + \sin u) \).Thus, the height simplifies to: \[ \cos(u + \pi/4) - \sin(u + \pi/4) = \sqrt{2}(-\sin u) \] The integral becomes: \[ V = \int_{0}^{\pi} -2\pi\sqrt{2}u\sin u \, du \]

Apply Symmetry and Evaluate

Since the height expression is odd and the limits are symmetric around \( u = \pi/2 \), we can apply symmetry and rewrite the integral:\[ V = 2\left(\int_{0}^{\pi/2} -2\pi\sqrt{2}u\sin u \, du\right) \].Integrate using integration by parts where \( dv = \sin u \, du \) and \( v = -\cos u \), and \( u \) is recycled. Carry out these calculations to find the final volume.

Simplify and Find the Volume

Use integration by parts to evaluate \( \int_{0}^{\pi/2} u\sin u \, du \). After evaluation, substitute back where needed and simplify to find:\[ V = 4\pi\left( -2\pi\sqrt{2}\right)\left( \frac{\pi}{4} - \frac{1}{2} \right) = 2\pi^2\sqrt{2} \].

Key concepts

Volume of Revolution

When you want to find the volume of a solid generated by revolving a region around an axis, the method of cylindrical shells is a fantastic tool. Imagine sweeping through the region one slice at a time. Each slice, or shell, is like a cylindrical surface. The shell's volume can be calculated, and when summed over the interval, it gives the full solid's volume.
The cylindrical shells method calculates volume using:

• Radius: This is the distance from the axis of rotation to the element of the region. In this problem, the radius is expressed as \( x - (-\pi/4) = x + \pi/4 \).
• Height: Corresponds to the difference in the functions defining the region's boundaries. Here, it is \( \cos x - \sin x \).
• Integral Expression: The objective is to express the integral as \( V = \int 2\pi \cdot (\text{radius}) \cdot (\text{height}) \ dx \).

This method simplifies multidimensional problems into more manageable ones by imagining the geometry of slicing the solid.

Trigonometric Integration

Trigonometric functions like sine and cosine show periodic behavior, which makes problems involving them interesting and sometimes challenging. When integrating expressions like these, applying trigonometric identities can simplify the work.
In our scenario, once we have the substitution \( u = x - \pi/4 \), the functions are transformed. The integration task then involves expressions like \( \cos(u + \pi/4) - \sin(u + \pi/4) \).

• Angle Addition Formulas: These are especially useful. For example, \( \cos(a + b) = \cos a \cdot \cos b - \sin a \cdot \sin b \).
• Resulting Expressions: Here, using the formula transforms the expression such that the difference simplifies to \( \sqrt{2}(-\sin u) \).

These identities simplify the integration process, making the entire computation more straightforward.

Symmetry in Integrals

Symmetry can be a powerful simplifier in integration problems. If a function is symmetric about a line or point, this characteristic can reduce the calculation load.
In our integral \( \int_{0}^{\pi} u \sin u \ du \), the symmetry comes into play because the function involved is odd within symmetric bounds. In simple terms, an odd function \( f(-x) = -f(x) \) integrated across an interval symmetric about the origin, or middle point here at \( u=\pi/2 \), will yield zero.
For this exercise, instead of computing the entire integral:

• Observe the integral bounds and function's behavior symmetry around any smart chosen midpoint.
• Utilize the symmetric properties, sometimes halving the integral and doubling it again for simplification, or rewriting using properties.

Thus, recognizing and exploiting symmetry makes the integral easier to manage, reducing complexity without altering the result.

Integration by Parts

Integration by parts is a method similar to the product rule for differentiation, useful in tackling integrals of products of functions. It's applied when standard methods might become cumbersome.
The formula, \( \int u \ dv = uv - \int v \ du \), helps reshuffle the terms to simplify certain integrals.

Steps for Using Integration by Parts:

• Choose \( u \) and \( dv \): For the term \( \int u \sin u \ du \), set \( u = u \) and \( dv = \sin u \ du \).
• Differentiation and Integration: Differentiate \( u \) to get \( du = 1 \ du \). Integrate \( dv \) to find \( v = -\cos u \).
• Re-substitute and Simplify: Insert these back into the integration by parts expression to simplify the result.

This approach reduces complicated integrals to more manageable ones repeatedly until the problem is solved, as we discovered when resolving \( V \) in our problem.