Problem 47 - Single Variable Calculus

Question

Find the volume of the solid generated by revolving the region enclosed by the graph of \(x^{1 / 2}+y^{1 / 2}=a^{1 / 2}\) and the coordinate axes about the \(x\) -axis.

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the region enclosed by the graph of \(x^{1 / 2}+y^{1 / 2}=a^{1 / 2}\) and the coordinate axes about the x-axis can be found using the disk method. The volume integral is given by: \(V = \pi \int_{0}^{a} \left(\frac{a^2 - x^2}{a + x - 4x^{1/2} (\frac{a^2 - x^2}{a + x - 4x^{1/2}y^{1/2}})^{1/2}}\right)^2 dx\) Due to the complexity of the integral, it is recommended to use a computer algebra system or software like WolframAlpha to evaluate it.

Step 1: Rewrite the equation

First, we need to isolate y in terms of x. The given equation is \(x^{1 / 2}+y^{1 / 2}=a^{1 / 2}\). To isolate y, follow these steps: Square both sides of the equation to remove the square roots: \((x^{1/2}+y^{1/2})^2=(a^{1/2})^2\) Multiply the left side of the equation: \(x+y+2(x^{1/2}y^{1/2})=a\) Move x to the other side in the equation: \(y = a - x - 2(x^{1/2}y^{1/2})\) Square both sides of the equation again: \(y^2 = (a - x - 2(x^{1/2}y^{1/2}))^2\) Expand the equation: \(y^2 = a^2 - 2ax + x^2 - 4(a-x)y^{1/2} + 4y\) Move all terms except the ones with a square root of y to the other side: \(y^2 - 4y = a^2 - 2ax + x^2 - 4(a -x)y^{1/2}\) Factor the left side: \(y(y - 4) = (a - x)(a + x - 4y^{1/2})\) Now, we have y isolated: \(y = \frac{a^2 - x^2}{a + x - 4x^{1/2}y^{1/2}}\)

Step 2: Set up the volume integral using the disk method

To find the volume of the solid generated by revolving the region about the x-axis, we will use the disk method. The volume can be found using the following integral: \(V = \pi \int_{0}^{a} [f(x)]^2 dx\) Where f(x) is the equation for y in terms of x. In this case, f(x) is: \(f(x) = \frac{a^2 - x^2}{a + x - 4x^{1/2}y^{1/2}}\) Therefore, the volume integral becomes: \(V = \pi \int_{0}^{a} \left(\frac{a^2 - x^2}{a + x - 4x^{1/2}y^{1/2}}\right)^2 dx\)

Step 3: Evaluate the integral

To evaluate the integral, substitute the equation for y back into the integral: \(V = \pi \int_{0}^{a} \left(\frac{a^2 - x^2}{a + x - 4x^{1/2} (\frac{a^2 - x^2}{a + x - 4x^{1/2}y^{1/2}})^{1/2}}\right)^2 dx\) This integral is highly complex and it is recommended to use a computer algebra system or a software like WolframAlpha to compute the definite integral. Once the integral is evaluated, you will have the volume of the solid generated by revolving the region enclosed by the graph of \(x^{1 / 2}+y^{1 / 2}=a^{1 / 2}\) and the coordinate axes about the x-axis.

Key Concepts

Disk Method

When faced with the challenge of calculating the volume of a solid of revolution, the disk method shines as an accessible approach. This technique involves slicing the solid into thin, disc-shaped pieces perpendicular to the axis of rotation.

Imagine a slice of a solid, where its edge traces a curve on a graph when revolving around an axis. The area of this disk is determined by the square of the radius – which is the function value at that point – and π, since the shape is circular. To get the total volume, we sum the volumes of these infinitesimal discs using a definite integral over the interval of revolution.

Specifically, the formula for the volume using the disk method is given by \[V = \pi \int_{a}^{b} [f(x)]^2 dx\], where \([f(x)]^2\) represents the area of the disk, and 'a' to 'b' is the range for the revolving region along the axis.

Definite Integral

A definite integral serves as a cornerstone in calculus, principally when dealing with problems that incorporate area, volume, and other concepts in spatial dimensions. It precisely captures the notion of net area under a curve, between the function graph and the x-axis, across the interval [a, b].

The fundamental theorem of calculus authorizes the use of antiderivatives to calculate definite integrals. Hence, evaluating \(\int_{a}^{b} f(x) dx\) results in F(b) - F(a), where F is the antiderivative of f. In the context of volume calculation, we use the definite integral to add up the infinitely many areas of disks, resulting in the total volume of a solid. Although integrating may be intricate, especially for complex functions, technology aids such as computer algebra systems can prove invaluable.

Volume Integral

Extending the idea of definite integrals, a volume integral can be considered a 3-dimensional counterpart that computes the volume under a surface within a given region. For solids of revolution, we simplify the process by revolving a 2D shape to get a 3D object and then take the integral of the cross-sectional area across the interval of rotation.

Using the disk method within the volume integral framework, we're integrating the cross-sectional area of the solid (i.e., the area of the disk) along the axis. This volume integral for solids of revolution is typically expressed as \(V = \pi \int_{a}^{b} [f(x)]^2 dx\), focusing solely on the radial dimension – as the function defines the radius of the disks at any given point along the axis.

Isolating Variables

Isolating variables is a technique used in algebra to manipulate equations so that one variable stands alone on one side of the equation. It's particularly useful in calculus when we need a function that exclusively expresses one variable in terms of others – typically y in terms of x or vice versa.

To isolate a variable, operations such as addition, subtraction, multiplication, division, and taking roots are performed systematically to both sides of the equation. These steps aim to eliminate other terms and leave the target variable by itself. When computing the volume of a solid of revolution, we frequently need to start by isolating the variable to avoid employing implicit functions, which are often more complicated to work with in integration.