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Chapter 5: Problem 29

Find the volume generated by rotating the area bounded by the graphs of each set of equations around the \(y\) -axis. $$ y=\sqrt{x^{2}+1}, x=0, x=3 $$

Short Answer

Expert verified
Volume = \(\pi \left( \frac{10 \sqrt{10}}{3} - \sqrt{10} + \frac{2}{3} \right)\) cubic units.

Step by step solution

01

Identify the region to be rotated

We have the curve given by the equation \(y = \sqrt{x^2 + 1}\) from \(x=0\) to \(x=3\). This curve represents a segment of a hyperbola that we want to rotate around the \(y\)-axis.
02

Set up the integral for volume using the disk method

In the disk method, the volume of the solid of revolution about the \(y\)-axis is given by\[V = \pi \int_{y_1}^{y_2} [g(y)]^2 \, dy\]where \([g(y)]^2\) is the radius of each disk perpendicular to the \(y\)-axis. In our case, solving for \(x\) gives \(x = \sqrt{y^2 - 1}\), and our bounds are from \(y = \sqrt{0^2 + 1} = 1\) to \(y = \sqrt{3^2 + 1} = \sqrt{10}\).
03

Transform the integral in terms of y

The integral for the volume becomes\[V = \pi \int_{1}^{\sqrt{10}} (y^2 - 1) \, dy\]This is obtained by computing \((\sqrt{y^2 - 1})^2 = y^2 - 1\) as the function describing the radius of the disks.
04

Compute the integral

Carry out the integral step-by-step:1. Integrate \(y^2\) with respect to \(y\):\(\int y^2 \, dy = \frac{y^3}{3}\)2. Integrate \(-1\) with respect to \(y\):\(\int -1 \, dy = -y\)Combined, we have:\[V = \pi \left[ \frac{y^3}{3} - y \right]_{1}^{\sqrt{10}} \]
05

Evaluate the integral at the bounds

Substitute \(y=\sqrt{10}\) and \(y=1\) into the integrated form:\[V = \pi \left( \frac{(\sqrt{10})^3}{3} - \sqrt{10} - \left( \frac{1^3}{3} - 1 \right) \right)\]Calculate each part:- \((\sqrt{10})^3 = 10^{3/2} = 10 \cdot \sqrt{10}\)- Evaluate: \[V = \pi \left( \frac{10 \sqrt{10}}{3} - \sqrt{10} - \left( \frac{1}{3} - 1 \right)\right)\]This simplifies further:
06

Final simplification

Simplify the expression to find the volume:Subtract and simplify each part:\[V = \pi \left( \frac{10 \sqrt{10}}{3} - \sqrt{10} + 1 - \frac{1}{3}\right) \]Combine like terms:\[V = \pi \left( \frac{10 \sqrt{10}}{3} - \sqrt{10} + \frac{2}{3} \right)\] Which gives us the final volume.

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Key Concepts

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

Disk Method
The disk method is a useful technique for finding the volume of a solid of revolution. When you want to find the volume generated by rotating a region around an axis, you imagine slicing the solid into thin, disk-like slices. Each slice is perpendicular to the axis of rotation.

Here's how it works:
  • You calculate the radius of each disk, which is the distance from the axis of rotation to the edge of the region being rotated.
  • For our example, when rotating around the \(y\)-axis, the radius of each disk depends on the \(x\)-value, expressed in terms of \(y\), as \(x = \sqrt{y^2 - 1}\).
  • The volume of each thin disk is approximately \(\pi \times \text{radius}^2 \times \Delta y\), where \(\Delta y\) is the thickness.
  • To find the total volume, you integrate these disks from the start to the end of the region in the \(y\)-direction.
This method helps to visualize and compute the volume of more complex shapes by breaking them down into simpler parts.
Solid of Revolution
A solid of revolution is a three-dimensional shape formed by rotating a two-dimensional region around an axis. These fascinating objects often resemble familiar items like cones, cylinders, or doughnuts, depending on the original region's shape.

To create a solid of revolution:
  • Pick a region in the plane that is bounded by one or more curves.
  • Decide on an axis of rotation. This could be the \(x\)-axis, \(y\)-axis, or any other line.
  • As the region rotates around the chosen axis, it sweeps out a three-dimensional shape.
  • For our exercise, the region bounded by \(y=\sqrt{x^2 + 1}\), \(x=0\), and \(x=3\) is rotated around the \(y\)-axis to form this solid.
Understanding the concept of a solid of revolution is crucial in solving problems related to volume, as it lays the foundation for applying techniques like the disk method.
Definite Integral
The definite integral is a fundamental tool in calculus. It allows us to compute the accumulated quantity, like area under a curve or volume of a solid, over an interval.

Here's how a definite integral works in the context of volume:
  • The integral sums up an infinite number of infinitesimally small quantities, like the volume of each disk in the disk method.
  • The bounds of the integral define the interval over which you conduct this summing process. For our volume calculation, the bounds were from \(y = 1\) to \(y = \sqrt{10}\).
  • The function inside the integral represents the quantity we're summing, such as the area of disks. In our solution, \([g(y)]^2 = y^2 - 1\) represents this.
  • By evaluating the definite integral with given bounds, you calculate the total volume of the solid.
Understanding definite integrals helps to tackle a wide range of problems in physics, engineering, and beyond, wherever accumulation is involved.

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Most popular questions from this chapter

The capitalized cost, \(c,\) of an asset over its lifetime is the total of the initial cost and the present value of all maintenance expenses that will occur in the future. It is computed with the formula $$ c=c_{0}+\int_{0}^{L} m(t) e^{-k t} d t $$ where \(c_{0}\) is the initial cost of the asset, \(L\) is the lifetime (in years), \(k\) is the interest rate (compounded continuously), and \(m(t)\) is the annual cost of maintenance. Find the capitalized cost under each set of assumptions. $$ \begin{array}{l} c_{0}=\$ 600,000, k=4 \% \\ m(t)=\$ 40,000+\$ 1000 e^{0.01 t}, L=40 \end{array} $$

Let \(x\) be a continuous random variable with a standard normal distribution. Using Table A, find each of the following. $$ P(-1.27 \leq x \leq-0.58) $$

(a) write a differential equation that models the situation, and (b) find the general solution. If an initial condition is given, find the particular solution. Recall that when \(y\) is directly proportional to \(x,\) we have \(y=k x\), and when \(y\) is inversely proportional to \(x,\) we have \(y=k / x,\) where \(k\) is the constant of proportionality. In these exercises, let \(k=1\). The rate of change of \(y\) with respect to \(x\) is directly proportional to the cube of \(y\).

Let \(x\) be a continuous random variable with a standard normal distribution. Using Table A, find each of the following. $$ P(-1.89 \leq x \leq 0.45) $$

Solve for \(y .\) \(y^{\prime}=3 x+x y ; \quad y=5\) when \(x=0\)
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