Question

An ellipse centered at the origin is described by the equation \(x^{2} / a^{2}+y^{2} / b^{2}=1 .\) If an ellipse \(R\) is revolved about either axis, the resulting solid is an ellipsoid. a. Find the volume of the ellipsoid generated when \(R\) is revolved about the \(x\) -axis (in terms of \(a\) and \(b\) ). b. Find the volume of the ellipsoid generated when \(R\) is revolved about the \(y\) -axis (in terms of \(a\) and \(b\) ). c. Should the results of parts (a) and (b) agree? Explain.

Short answer

Explain. Answer: No, the volumes should not agree. This is because the resulting ellipsoids have different shapes depending on which axis the ellipse is revolved around. When the ellipse is revolved around the x-axis, the major axis of the ellipse becomes the minor axis of the ellipsoid, and when we revolve the ellipse around the y-axis, the minor axis of the ellipse becomes the major axis of the ellipsoid. Thus, the ellipsoids have different shapes and consequently different volumes.

Step-by-step solution

Part a: Find the volume when revolving around the x-axis

To find the volume of the ellipsoid generated when ellipse R is revolved around the x-axis, we can use the disk method. The disk method involves integrating the area of each infinitesimally small disk generated by revolving a small interval dy around the x-axis. The area of each disk will be the cross-sectional area of the ellipsoid at that point, and we will sum these areas along the ellipse's height. The equation of the ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). We need to rearrange this equation to solve for y in terms of x: \(y^2 = b^2(1 - \frac{x^2}{a^2})\), and then \(y=b\sqrt{1 - \frac{x^2}{a^2}}\). The area of the cross-section at a given x-value will be \(A(x)=\pi y^2=\pi b^2 \left(1-\frac{x^2}{a^2}\right)\). To find the volume, we integrate this expression with respect to x from -a to a: \(V = \int_{-a}^{a} A(x)dx = \int_{-a}^{a} \pi b^2 \left(1-\frac{x^2}{a^2}\right) dx\)

Evaluate the integral

Let's evaluate the integral step by step: \(V = \pi b^2 \int_{-a}^{a} \left(1-\frac{x^2}{a^2}\right) dx = \pi b^2 \left[x - \frac{x^3}{3a^2}\right]_{-a}^{a}\) \(V = \pi b^2 \left[ \left(a - \frac{a^3}{3a^2}\right) - \left(-a - \frac{(-a)^3}{3a^2}\right) \right]\) \(V = \pi b^2 \left[ a - \frac{a}{3} + a - \frac{a}{3} \right] = \pi b^2 \left[\frac{4}{3}a\right]\) \(V = \boxed{\frac{4}{3} \pi a b^2}\) So, when the ellipse is revolved about the x-axis, the volume of the resulting ellipsoid is \(\frac{4}{3}\pi a b^2.\)

Part b: Find the volume when revolving around the y-axis

Now, to find the volume of the ellipsoid generated when ellipse R is revolved around the y-axis, we again use the disk method. This time, the area of each disk will be the cross-sectional area of the ellipsoid at a given y-value. We need to rearrange the ellipse equation to solve for x in terms of y: \(x^2 = a^2(1 - \frac{y^2}{b^2})\), and then \(x=a\sqrt{1 - \frac{y^2}{b^2}}\). The area of the cross section at a given y-value will be \(A(y)=\pi x^2=\pi a^2 \left(1-\frac{y^2}{b^2}\right)\). To find the volume, we integrate this expression with respect to y from -b to b: \(V = \int_{-b}^{b} A(y)dy = \int_{-b}^{b} \pi a^2 \left(1-\frac{y^2}{b^2}\right) dy\)

Evaluate the integral

Let's evaluate the integral step by step: \(V = \pi a^2 \int_{-b}^{b} \left(1-\frac{y^2}{b^2}\right) dy = \pi a^2 \left[y - \frac{y^3}{3b^2}\right]_{-b}^{b}\) \(V = \pi a^2 \left[ \left(b - \frac{b^3}{3b^2}\right) - \left(-b - \frac{(-b)^3}{3b^2}\right) \right]\) \(V = \pi a^2 \left[ b - \frac{b}{3} + b - \frac{b}{3} \right] = \pi a^2 \left[\frac{4}{3}b\right]\) \(V = \boxed{\frac{4}{3} \pi a^2 b}\) So, when the ellipse is revolved about the y-axis, the volume of the resulting ellipsoid is \(\frac{4}{3}\pi a^2 b.\)

Part c: Should the results agree?

In this part of the problem, we are asked whether the volumes of the ellipsoids produced in parts (a) and (b) should agree or not. We found that when the ellipse is revolved around the x-axis, the volume is \(\frac{4}{3}\pi a b^2\), and when the ellipse is revolved around the y-axis, the volume is \(\frac{4}{3}\pi a^2 b\). Although these two expressions may appear similar, they are not equal since the exponent of a and b is different in each case. These results should not agree, because the two ellipsoids produced are not the same. When we revolve the ellipse around the x-axis, the major axis of the ellipse becomes the minor axis of the ellipsoid, and when we revolve the ellipse around the y-axis, the minor axis of the ellipse becomes the major axis of the ellipsoid. Thus, depending on which axis we choose to revolve the ellipse, the resulting ellipsoids have different shapes and consequently different volumes.

Key concepts

Disk Method

The Disk Method is a powerful technique in calculus used to find the volume of a solid of revolution. Imagine slicing the object into many thin disks, similar to pancakes stacked one on top of another.
Each disk is a cross-section of the object, with a circular shape. To use this method, you revolve a shape around an axis which in turn forms these disks.

• The radius of each disk depends on the function creating the solid.
• The thickness of each disk is an infinitesimally small change, denoted as \(dx\) or \(dy\), depending on the axis of revolution.

Revolution around the x-axis involves integrating the area of these disks with respect to \(x\), while revolutions around the y-axis require integration with respect to \(y\). For ellipsoids, the equation of the boundary determines the varying radii for these disks.
Applying the disk method helps us not only to comprehend the formation of complex solids but also their volumes, connecting geometry with calculus seamlessly.

Integration in Calculus

Integration is a core concept in calculus, crucial for calculating the volume of solids, like an ellipsoid, or areas under curves. It's the process of summing an infinite number of infinitesimally small quantities.
In our context, integration lets us add up all the small volumes of disks to get the total volume of the solid.
This is expressed in an integral:

• Set up the integral according to the axis of revolution.
• Determine limits of integration based on where the solid starts and ends on that axis.
• Use the function to define the radius and subsequently the area of the disks.

The formula for the volume of the ellipsoid involves integrating these areas over the specified range. This mathematical tool makes it possible to find exact values, dealing with otherwise impossible calculations through simple summation.

Revolution of Solids

Revolution of solids refers to creating a 3D object by rotating a 2D shape around a line called the axis of revolution. This method gives rise to solids like cylinders, spheres, and ellipsoids, depending on the original shape.
When we rotate an ellipse around the x-axis or y-axis, it results in an ellipsoid, a type of solid with both rotational symmetry and distinct axes lengths.
For an ellipse described by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), revolving around the x-axis, the ellipse stretches out to create an ellipsoid with the volume formula \({\frac{4}{3}\pi a b^2}\).

• Revolving around different axes changes the dimensions of the ellipsoid.
• Each axis forms a different solid, evident by their various volume equations.

Understanding the revolution of solids explicates how movement in space affects volume and shape, providing insight into the geometric properties of solids formed by rotation.