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Chapter 6: Problem 51

The region under the graph of \(y=\sqrt{\cos ^{-1} x}\) on the interval \([0,1]\) is revolved about the \(x\) -axis. Find the volume of the solid generated.

Short Answer

Expert verified
The volume of the solid generated by revolving the region under the graph of \(y=\sqrt{\cos ^{-1} x}\) on the interval \([0,1]\) about the x-axis is approximately \(\frac{蟺^2}{4}\).

Step by step solution

01

Identify the function and the interval

The function is given as \(y=\sqrt{\cos ^{-1} x}\) and the interval is given as \([0,1]\).
02

Set up the integral for the disk method

The disk method involves integrating the square of the function with respect to x multiplied by the constant factor \(蟺\). So the integral would look like: \(V = \int_a^b 蟺y^2\,dx\) In our case, the function is \(y=\sqrt{\cos ^{-1} x}\). Then, the integral becomes: \(V = \int_0^1 蟺(\sqrt{ \cos ^{-1} x})^2\,dx\)
03

Simplify the integral

Squaring the function inside the integral, we get: \(V = \int_0^1 蟺(\cos ^{-1} x)\,dx\) Now, we need to solve this integral.
04

Integrate with respect to x

To solve the integral, we will perform integration by parts. Let: \(u = \cos ^{-1} x \Rightarrow du = -\frac{1}{\sqrt{1-x^2}}\,dx\) \(dv = dx \Rightarrow v = x\) Now, using the integration by parts formula, \(\int u\,dv = uv - \int v\,du\), we have: \(V = 蟺\Bigg[x\cos ^{-1} x \Bigg|_0^1 - \int_0^1 \frac{-x}{\sqrt{1-x^2}}\,dx\Bigg]\)
05

Solve the remaining integral

The remaining integral is \(\int_0^1 \frac{x}{\sqrt{1-x^2}}\,dx\). To solve this, we can use the substitution: \(w = 1 - x^2 \Rightarrow dw = -2x\,dx\) The integral becomes: \(\int \frac{-1}{2} \frac{dw}{\sqrt{w}}\) Now, we can easily integrate this expression: \(\int \frac{-1}{2} \frac{dw}{\sqrt{w}} = -\frac{1}{2} \int w^{-\frac{1}{2}}\,dw = -\frac{1}{2}\Bigg[ 2w^{\frac{1}{2}}\Bigg] = -\sqrt{w}\) Now, substitute back \(w = 1 - x^2\): \(-\sqrt{1 - x^2}\) So the original integral can be written as: \(V = 蟺\Bigg[x\cos ^{-1} x \Bigg|_0^1 + \int_0^1\sqrt{1-x^2}\,dx \Bigg]\)
06

Evaluate the definite integral and calculate the volume

To find the volume V, we need to evaluate the definite integral: \(V = 蟺\Bigg[\bigg(1\cdot \cos ^{-1} 1\bigg) - \bigg(0\cdot \cos ^{-1} 0\bigg) + \int_0^1\sqrt{1-x^2}\,dx \Bigg]\) We know that: \(\cos^{-1}(1) = 0\) and \(\cos^{-1}(0) = \frac{蟺}{2}\) Therefore: \(V = 蟺\Bigg[0 - 0 + \int_0^1\sqrt{1-x^2}\,dx \Bigg]\) Finding the exact value for the integral \(\int_0^1\sqrt{1-x^2}\,dx\) is quite involved, but we can approximate its value using numerical methods (such as the Simpson's rule) or using a calculator, which gives approximately: \(\int_0^1\sqrt{1-x^2}\,dx \approx \frac{蟺}{4}\) Thus, the volume of the solid generated is: \(V = 蟺\Bigg[\frac{蟺}{4}\Bigg] = \frac{蟺^2}{4}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Parts
Integration by Parts is a powerful technique in calculus used to integrate the product of two functions. The core idea is to transform a difficult integral into a simpler one. It鈥檚 based on the product rule for differentiation and is represented by the formula:\[\int u \, dv = uv - \int v \, du\]Here is how you apply it:
  • Identify two parts of the integrand: the function to differentiate \( u \), and the function to integrate \( dv \).
  • Differentiate \( u \) to get \( du \).
  • Integrate \( dv \) to get \( v \).
  • Apply the integration by parts formula to find the integral.
In our exercise, we integrated the function \( y = \sqrt{\cos^{-1} x} \), which required setting \( u = \cos^{-1} x \) and \( dv = dx \). This clever choice allowed us to simplify the integral and ultimately find the volume of the solid.
Definite Integral
A definite integral is a fundamental concept in calculus that not only helps to find the area under a curve but also the accumulation of quantities. When you compute a definite integral, you are essentially evaluating the integral over a specific interval \([a, b]\). It's represented as:\[\int_a^b f(x) \, dx\]To evaluate a definite integral:
  • Find the antiderivative of the function, also known as the indefinite integral.
  • Use the limits \( a \) and \( b \) to find the difference between the values of the antiderivative at these points.
For our problem, the definite integral \( \int_0^1 \pi (\cos^{-1} x) \, dx \) was used to determine the volume of the solid. We used the antiderivative rules and integration techniques to solve it over the interval \([0, 1]\). This process ultimately gives us precise calculations on physical quantities, such as volume in this case.
Volume of Revolution
The concept of the volume of revolution involves finding the volume of a solid formed by rotating a region around an axis. The Disk Method is a common technique used to solve these problems. Here鈥檚 how it works:
  • Identify the function that generates the region to be revolved.
  • Set up the integral using the area of a disk, which is \( \pi y^2 \) for a revolution around the x-axis.
  • Determine the limits of integration based on the interval of the region.
In our specific problem, the function \( y = \sqrt{\cos^{-1} x} \) defined the region. By revolving this region around the x-axis from \( x = 0 \) to \( x = 1 \), we used the Disk Method to calculate the volume of the resulting solid. This approach, therefore, helps in translating geometrical shapes into computable volumes which are essential in real-world applications like engineering and physics.

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Most popular questions from this chapter

Prove that if \(m\) and \(n\) are positive integers, then $$ \int_{-\pi}^{\pi} \sin m x \sin n x d x=\left\\{\begin{array}{ll} 0 & \text { if } m \neq n \\ \pi & \text { if } m=n \end{array}\right. $$

Observe that \(\int_{0}^{\infty} e^{-x^{2}} d x=\int_{0}^{4} e^{-x^{2}} d x+\int_{4}^{\infty} e^{-x^{2}} d x\). a. Show that \(\int_{4}^{\infty} e^{-x^{2}} d x \leq 10^{-7}\) so that \(\int_{0}^{\infty} e^{-x^{2}} d x \approx \int_{0}^{4} e^{-x^{2}} d x\) b. Use a calculator or computer to obtain an estimate for $\int_{0}^{\infty} e^{-x^{2}} d x.

Find or evaluate the integral. $$ \int \frac{d x}{x \sqrt{1+(\ln x)^{2}}} $$

The graph \(C\) of the function $$ y=\frac{P \alpha}{2 k} e^{-\alpha|x|}(\cos \alpha x+\sin \alpha|x|) $$ where \(\alpha\) and \(k\) are constants, gives the shape of a beam of infinite length lying on an elastic foundation and acted upon by a concentrated load \(P\) applied to the beam at the origin. Before application of the force, the beam lies on the \(x\) -axis. Find the potential energy of elastic deformation \(W\) using the formula $$ W=E e \int_{0}^{\infty}\left(y^{\prime \prime}\right)^{2} d x $$ where \(E\) and \(e\) are constants. Note: This model provides a good approximation in working with long beams.

Plot the graph of \(f(x)=e^{-x} \sin (\pi x / 2)\) for \(0 \leq x \leq 2\). Then use a calculator or computer to approximate the volume of the solid generated by revolving the region under the graph of \(f\) on \([0,2]\) about (a) the \(x\) -axis and (b) the \(y\) -axis.
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