Problem 54
Question
Volume of an Ellipsoid Consider the plane region bounded by the graph of $$\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1$$ where \(a>0\) and \(b>0 .\) Show that the volume of the ellipsoid formed when this region is revolved about the \(y\) -axis is $$\frac{4}{3} \pi a^{2} b$$ What is the volume when the region is revolved about the \(x\) -axis?
Step-by-Step Solution
Verified Answer
The volume of the ellipsoid formed when the region is revolved about the y-axis is \(\frac{4}{3} \pi a^{2}b\), and when revolved about the x-axis is \(\frac{4}{3} \pi ab^{2}\).
1Step 1: Calculate volume with revolve around y-axis
Rewrite the ellipse equation \(x = a \sqrt{1-\left(\frac{y}{b}\right)^{2}}\). Substitute it to the formula of the volume of a solid of revolution. \(V_{y} = \pi \int_{-b}^{b} \left(a \sqrt{1-\left(\frac{y}{b}\right)^{2}}\right)^{2} dy\). This simplifies to \(V_{y} = \pi \int_{-b}^{b} a^{2} \left(1-\left(\frac{y}{b}\right)^{2}\right) dy\). Integrating yields \(V_{y} =\frac{4}{3} \pi a^{2}b\).
2Step 2: Calculate volume with revolve around x-axis
Rewrite the ellipse equation \(y = b \sqrt{1-\left(\frac{x}{a}\right)^{2}}\). Substitute it to the formula of the volume of a solid of revolution. \(V_{x} = \pi \int_{-a}^{a} \left(b \sqrt{1-\left(\frac{x}{a}\right)^{2}}\right)^{2} dx\). This simplifies to \(V_{x} = \pi \int_{-a}^{a} b^{2} \left(1-\left(\frac{x}{a}\right)^{2}\right) dx\). Integrating yields \(V_{x} =\frac{4}{3} \pi ab^{2}\).
Key Concepts
Solids of RevolutionIntegrationCalculusVolume Calculation
Solids of Revolution
A solid of revolution is a three-dimensional object created by rotating a two-dimensional shape around a straight line (axis of rotation) that lies on the same plane as the shape. For example, when a semicircle rotates around its diameter, it forms a sphere, and when a rectangle rotates around one of its sides, it forms a cylinder.
The idea of a solid of revolution is fundamental in the field of calculus, especially when trying to determine the volume of an object whose cross-section is known. In our exercise, the cross-section of an ellipsoid is given, and we must find its volume when rotated around both the x-axis and y-axis.
To visualize it, imagine a flat ellipse lying on a plane. If we rotate the ellipse around one of its principal axes, it sweeps out the shape of an ellipsoid. Each slice of the ellipsoid perpendicular to the axis of rotation is a circle whose radius varies depending on the position along the axis.
The idea of a solid of revolution is fundamental in the field of calculus, especially when trying to determine the volume of an object whose cross-section is known. In our exercise, the cross-section of an ellipsoid is given, and we must find its volume when rotated around both the x-axis and y-axis.
To visualize it, imagine a flat ellipse lying on a plane. If we rotate the ellipse around one of its principal axes, it sweeps out the shape of an ellipsoid. Each slice of the ellipsoid perpendicular to the axis of rotation is a circle whose radius varies depending on the position along the axis.
Integration
Integration is a core operation in calculus that, among other things, allows us to calculate areas, volumes, and lengths. When faced with a problem involving solids of revolution, integration comes into play as we work out the volume by summing an infinite number of infinitesimally thin disks or washers that constitute the solid.
In this case, for the ellipsoid formed by rotating the region around the y-axis, we evaluate the integral \(V_{y} = \pi \int_{-b}^{b} a^{2} \left(1-\left(\frac{y}{b}\right)^{2}\right) dy\). This integral represents the sum of the areas of circular disks stacked along the y-axis from \(y = -b\) to \(y = b\).
The integration of the function gives us the exact volume of the ellipsoid, which can be calculated by hand or with the aid of computational tools. The key is to set up the integral correctly by considering the shape and size of each disk at a given position along the axis.
In this case, for the ellipsoid formed by rotating the region around the y-axis, we evaluate the integral \(V_{y} = \pi \int_{-b}^{b} a^{2} \left(1-\left(\frac{y}{b}\right)^{2}\right) dy\). This integral represents the sum of the areas of circular disks stacked along the y-axis from \(y = -b\) to \(y = b\).
The integration of the function gives us the exact volume of the ellipsoid, which can be calculated by hand or with the aid of computational tools. The key is to set up the integral correctly by considering the shape and size of each disk at a given position along the axis.
Calculus
Calculus is a branch of mathematics that studies how things change. It's a tool that allows us to investigate the properties of functions and their graphs. Calculus is divided into two main parts: differential calculus and integral calculus.
Differential calculus focuses on the rate of change (derivatives), while integral calculus is about accumulation (integrals). Together, they form the basis for solving many problems involving motion, areas, volumes, and more. In our exercise, integral calculus is utilized to find the volume of an ellipsoid by integrating along the axis of revolution.
Understanding calculus concepts like limits, derivatives, and integrals are crucial for solving real-world problems. In the case of our ellipsoid, calculus enables us to transition from a two-dimensional ellipse to a three-dimensional volume through the process of rotation and integration.
Differential calculus focuses on the rate of change (derivatives), while integral calculus is about accumulation (integrals). Together, they form the basis for solving many problems involving motion, areas, volumes, and more. In our exercise, integral calculus is utilized to find the volume of an ellipsoid by integrating along the axis of revolution.
Understanding calculus concepts like limits, derivatives, and integrals are crucial for solving real-world problems. In the case of our ellipsoid, calculus enables us to transition from a two-dimensional ellipse to a three-dimensional volume through the process of rotation and integration.
Volume Calculation
Volume calculation in calculus is an application of integral calculus that deals with finding the three-dimensional space occupied by a shape. For solids of revolution, this involves slicing the solid into thin pieces, finding the area of each slice, and then summing these areas along the axis of rotation.
The volume of an ellipsoid, when rotated around the y-axis, is represented by the integral \(V_{y} = \pi \int_{-b}^{b} a^{2} \left(1-\left(\frac{y}{b}\right)^{2}\right) dy\), which simplifies to \(V_{y} =\frac{4}{3} \pi a^{2} b\). When rotated around the x-axis, we use a similar approach, resulting in the volume \(V_{x} =\frac{4}{3} \pi ab^{2}\).
These formulae resemble that of a sphere but are adjusted for the ellipsoidal shape. For complex shapes, computational methods or calculator functionalities might be necessary to evaluate the integrals for volume calculation.
The volume of an ellipsoid, when rotated around the y-axis, is represented by the integral \(V_{y} = \pi \int_{-b}^{b} a^{2} \left(1-\left(\frac{y}{b}\right)^{2}\right) dy\), which simplifies to \(V_{y} =\frac{4}{3} \pi a^{2} b\). When rotated around the x-axis, we use a similar approach, resulting in the volume \(V_{x} =\frac{4}{3} \pi ab^{2}\).
These formulae resemble that of a sphere but are adjusted for the ellipsoidal shape. For complex shapes, computational methods or calculator functionalities might be necessary to evaluate the integrals for volume calculation.
Other exercises in this chapter
Problem 53
The manufacturer of glass for a window in a conversion van needs to approximate its center of mass. A coordinate system is superimposed on a prototype of the gl
View solution Problem 54
Verifying a Formula (a) Given a circular sector with radius \(L\) and central angle \(\theta\) (see figure), show that the area of the sector is given by $$S=\f
View solution Problem 54
Integration as an Accumulation Process In Exercises \(51-54\) , find the accumulation function \(F\) . Then evaluate \(F\) at each value of the independent vari
View solution Problem 55
Lateral Surface Area of a Cone A right circular cone is generated by revolving the region bounded by \(y=3 x / 4\) , \(y=3,\) and \(x=0\) about the \(y\) -axis.
View solution