Question

Volume Find the volume of the solid obtained by revolving the region bounded by the curve \(y=\frac{1}{1-\sin x}\) on \([0, \pi / 4]\) about the \(x\) -axis.

Short answer

Question: Find the volume of the solid generated by revolving the curve \(y = \frac{1}{1-\sin x}\) for \(0 \le x \le \frac{\pi}{4}\) around the x-axis. Answer: The volume of the solid is \(V = \pi(1 - \sqrt{2})\).

Step-by-step solution

Set up the volume formula

Plug the function \(f(x) = \frac{1}{1-\sin x}\) and limits of integration into the volume formula: \(V = \pi \int_0^{\frac{\pi}{4}} \left(\frac{1}{1-\sin x}\right)^2 dx\)

Simplify the integral

Simplify the integral as: \(V = \pi \int_0^{\frac{\pi}{4}} \frac{1}{(1-\sin x)^2} dx\)

Perform integration

To evaluate the integral, we use the substitution method. Let \(u = 1 - \sin x\). Then, \(-du = \cos x dx\). Since \(\sin 0 = 0\) and \(\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}\), our limits of integration for u will become \(1\) and \(\frac{\sqrt{2}}{2}\). Thus, the integral becomes: \(V = \pi \int_1^{\frac{\sqrt{2}}{2}} \frac{-du}{u^2 \cos x} = -\pi \int_1^{\frac{\sqrt{2}}{2}} \frac{du}{u^2}\) Now, we perform the integration: \(V = -\pi \left[ \frac{1}{u} \right]_1^\frac{\sqrt{2}}{2} = \pi \left(\frac{1}{1} - \frac{1}{\frac{\sqrt{2}}{2}}\right) = \pi(1 - \sqrt{2})\)

Final Answer

The volume of the solid is \(V = \pi(1 - \sqrt{2})\).

Key concepts

Integration by Substitution

Integration by substitution is a vital technique in calculus used to simplify complex integrals. Think of it as a way to change the variable of integration to make the problem easier to solve.
For example, in the given problem, our original function for the volume formula involves \(y=\frac{1}{1-\sin x}\). The integral itself may be intimidating at first glance.
However, by making a clever substitution where \(u=1-\sin x\), we change the integral into a form we can more easily work with.

• In this context, we also identify \(-du=\cos x \ dx\) due to the derivative of \(\sin x\), which assists in transforming the \(dx\) to \(du\).
• This substitution technique rights the integral into an approachable \(-\int \frac{1}{u^2} \, du\) format.

Substitution effectively reorients the problem: the same boundaries but now in terms of \(u\). This method is fundamental when dealing with complex integrals like those involved in finding volumes of revolution.

Volume of Solid of Revolution

When we talk about the volume of a solid of revolution, it refers to a 3D shape created by rotating a 2D region around an axis. Think of twirling a flat shape like a pancake on a stick; what you get is a delightful doughnut-like geometry.
In the problem at hand, we revolve the region under the curve \(y=\frac{1}{1-\sin x}\) from \(x = 0\) to \(x = \pi/4\) around the \(x\)-axis.
The standard technique to find such volumes is the disk method, capturing the idea of breaking the solid into a stack of thin disks.
The volume of each disk is \(\pi \, [\text{radius}]^2 \, \Delta x\).

• The "radius" at any point of the curve is defined by the function value, here \(\frac{1}{1-\sin x}\).
• When accumulated over the interval \[0, \pi/4\], we find the total volume by integrating, using the definite integral.

This problem demonstrates a unique rotation that crafts beautiful and complex geometries, reminding us yet again of calculus' fascinating capabilities.

Definite Integration

Definite integration is the process of computing the integral of a function over a specific interval, providing the exact accumulation of quantities and not just a general form.
In this exercise, by using the definite integral \(\pi \int_0^{\frac{\pi}{4}} \left(\frac{1}{1-\sin x}\right)^2 dx\), we're determining not just an area, but a volume.
The bounds \(0\) to \(\pi/4\) signify the specific region along the \(x\)-axis we assess the function over.
Here are some highlights of why definite integrals matter:

• They help find exact total quantities—be it area, volume, or mass—between specified bounds.
• They provide solutions that directly answer physical real-world questions.
• This operation turns an infinite sum of infinitesimals into a tangible answer.

As we compute, the value \(V = \pi(1 - \sqrt{2})\) emerges, a clear testament to the power of definite integration in evaluating complex shapes and regions with precision.