Problem 43
Question
Find the volume of the torus generated by revolving the circle \((x-2)^{2}+y^{2}=1\) about the \(y\) -axis.
Step-by-Step Solution
Verified Answer
The volume of the torus is calculated using circular cross-sections revolved around the \(y\)-axis.
1Step 1: Identify the Elements of the Circle
The given circle equation is \((x-2)^2 + y^2 = 1\). This is a circle centered at \((2, 0)\) with a radius \(r = 1\).
2Step 2: Set Up the Volume Integral Using the Washer Method
To find the volume of the torus, we revolve the circle around the \(y\)-axis. The volume \(V\) is found using the washer method:\[V = 2\pi \int_{-1}^{1} (2)^2 - (2 - \sqrt{1 - y^2})^2 \, dy\]
3Step 3: Simplify the Expression Inside the Integral
Simplify the expression inside the integral:\[(2)^2 - (2 - \sqrt{1 - y^2})^2 = 4 - [(4 - 4\sqrt{1-y^2} + (1-y^2))] \= 4y^2 - 4\sqrt{1-y^2}\]
4Step 4: Evaluate the Integral
Evaluate each part separately if needed or through substitution. Notice this involves trigonometric substitution:\[x = \sin(\theta)\] implies \[dx = \cos(\theta) d\theta\], then process the integral accordingly. Substitution or numerical integration may be needed for detailed steps.
5Step 5: Calculate the Volume
Calculate the entire expression to get the volume of the torus. Complete the integration and multiply by any constants. This might require solving a definite integral for verification.
Key Concepts
Washer MethodRevolution about Y-AxisTrigonometric SubstitutionDefinite Integral
Washer Method
The washer method is a popular technique used to find the volume of a solid of revolution. It is especially useful when finding the volume of an object with a hole in the middle, like a washer or a torus.
- In essence, you're dealing with a series of concentric disks, which are stacked upon one another to fill the space of the solid.
- The washer method involves subtracting the volume of the inner disk from the outer disk at each point.
Revolution about Y-Axis
When you revolve a shape around an axis, it creates a 3D object. In this case, the exercise involves revolving a circle around the y-axis to form a torus.
- Revolving around the y-axis means every point on the circle describes a circular path, forming a toroidal shape.
- The revolution about the y-axis involves integrating in terms of y, which simplifies the mathematical manipulation when dealing with circles or curves symmetrical along the x-axis.
Trigonometric Substitution
Trigonometric substitution is a technique used to simplify integrals involving square roots. It leverages trigonometric identities to make the integration process more straightforward.
- This is especially useful when your integrand contains expressions like \(\sqrt{1-y^2}\), as it resembles the Pythagorean identity \(1 = \sin^2(\theta) + \cos^2(\theta)\).
- By setting \(y = \sin(\theta)\), we transform the square root into something manageable.
Definite Integral
The definite integral is a fundamental concept in calculus used to find the total accumulation of a quantity, such as area or volume, over an interval.
- In our context, it computes the total volume of the torus by integrating along the bounds of the chosen variable, y, from \(-1\) to \(1\).
- Definite integrals consider the limits of integration, unlike indefinite integrals which represent a family of antiderivatives.
Other exercises in this chapter
Problem 42
Use a theorem of Pappus to find the volume generated by revolving about the line \(x=5\) the triangular region bounded by the coordinate axes and the line \(2 x
View solution Problem 42
In Exercises \(39-42\) , find the volume of the solid generated by revolving each region about the \(y\) -axis. The region in the first quadrant bounded on the
View solution Problem 43
In Exercises 43 and \(44,\) find the volume of the solid generated by revolving each region about the given axis. The region in the first quadrant bounded above
View solution Problem 45
Use the Second Theorem of Pappus and the fact that the surface area of a sphere of radius \(a\) is 4\(\pi a^{2}\) to find the centroid of the semicircle \(y=\sq
View solution