Problem 34 - Calculus Early Transcendentals

Question

If the top half of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is revolved about the \(x\) -axis, the result is an ellipsoid whose axis along the \(x\) -axis has length \(2 a,\) whose axis along the \(y\) -axis has length \(2 b,\) and whose axis perpendicular to the \(x y\) -plane has length \(2 b .\) We assume that \(0

Step-by-Step Solution

Answer

Question: Show that the surface area of an ellipsoid formed by revolving the top half of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) about the x-axis is equal to the surface area of a sphere when \(a=b\). Answer: When \(a=b\), the surface area of the ellipsoid is equal to \(4\pi a^2\), which is the surface area of a sphere of radius \(a\).

Step 1: Derive the surface area formula for the ellipsoid

Using the given ellipse equation \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), we can isolate \(y\) to obtain \(y=b\sqrt{1-\frac{x^2}{a^2}}\) since we are only considering the upper half of the ellipse. Now, we can find the surface area \((S)\) by using the formula for the surface area of revolution: $$ S = 2\pi \int_{0}^{a} y \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx $$ Differentiate \(y\) with respect to \(x\) to get \(\frac{dy}{dx} = -\frac{bx}{a^2\sqrt{1-\frac{x^2}{a^2}}}\) (by using the chain rule). Now, square \(\frac{dy}{dx}\) and simplify to obtain \(\left(\frac{dy}{dx}\right)^2 = \frac{b^2x^2}{a^4\left(1-\frac{x^2}{a^2}\right)}\). Add 1 to this expression and then take the square root, yielding: $$\sqrt{1+\left(\frac{dy}{dx}\right)^2} = \sqrt{\frac{a^4-b^2x^2}{a^4\left(1-\frac{x^2}{a^2}\right)}} = \frac{1}{a}\sqrt{a^2-c^2x^2}$$ where \(c^2=1-\frac{b^2}{a^2}\). Now, we can plug this expression back into the surface area integral: $$ S = 2\pi \int_{0}^{a} b\sqrt{1-\frac{x^2}{a^2}} \cdot \frac{1}{a}\sqrt{a^2-c^2x^2} dx = \frac{4\pi b}{a} \int_{0}^{a} \sqrt{a^2-c^2x^2} dx $$ #b. Change of Variables#

Step 2: Perform the change of variables in the surface area integral

Next, we perform the change of variables \(u=cx\) where \(c^2 = 1 - \frac{b^2}{a^2}\). This means that \(du = c dx\). Also, when \(x=0\), we get \(u=0\), and when \(x=a\), we get \(u=\sqrt{a^2-b^2}\). We can rewrite the surface area integral as: $$ S = \frac{4\pi b}{a} \int_{0}^{\sqrt{a^2-b^2}} \sqrt{a^2-\frac{u^2}{c^2}} \frac{du}{c} = \frac{4\pi b}{\sqrt{a^2-b^2}} \int_{0}^{\sqrt{a^2-b^2}} \sqrt{a^2-u^2} du $$ #c. Apply Integral Formula#

Step 3: Apply the given integral formula to find the surface area of the ellipsoid

Use the given integral formula to calculate the surface area: $$ \int \sqrt{a^2-u^2} du = \frac{1}{2}\left[u\sqrt{a^2-u^2}+a^2\sin^{-1}\frac{u}{a}\right]+C $$ By substituting the limits of integration, we can find the surface area of the ellipsoid: $$ S = 2\pi b \left( b+\frac{a^2}{\sqrt{a^2-b^2}}\sin^{-1}\frac{\sqrt{a^2-b^2}}{a} \right) $$ #d. Units of S#

Step 4: Find the units of the surface area

Given that \(a\) and \(b\) have units of length (e.g. meters), the surface area formula has the following units: $$ S = [2\pi b] \cdot \left[[b] + \left[\frac{a^2}{\sqrt{a^2-b^2}}\right][\sin^{-1}\frac{\sqrt{a^2-b^2}}{a}]\right] $$ Since trigonometric functions are unitless and all the lengths inside the square root are squared, the units inside the square brackets will cancel out. This means that the units of \(S\) are the same as \(2\pi b\), which are \((\text{length})^2\). Therefore, the surface area \(S\) has units of square meters. #e. Surface Area when a=b#

Step 5: Show that the surface area of the ellipsoid is equal to the surface area of a sphere when \(a=b\)

When \(a=b\), we can rewrite the ellipse equation as \(\frac{x^2}{a^2}+\frac{y^2}{a^2}=1\), which represents a circle of radius \(a\). Let's use the surface area formula derived in part (a) to find the surface area when \(a=b\): $$ S=\frac{4\pi b}{a} \int_{0}^{a} \sqrt{a^{2}-c^{2} x^{2}} dx $$ Since \(a=b\), \(c^2=1-\frac{b^2}{a^2}=0\), and the integral becomes: $$ S=\frac{4\pi b}{a} \int_{0}^{a} \sqrt{a^{2}-0 \cdot x^{2}} dx = \frac{4\pi b}{a} \int_{0}^{a} \sqrt{a^{2}} dx = 4\pi a^2$$ which is the surface area of a sphere of radius \(a\).

Key Concepts

Ellipsoid Revolution

When you revolve the upper half of an ellipse around the x-axis, you create a three-dimensional shape called an ellipsoid of revolution. This geometric shape has three distinctive axes measurements:

• The axis along the x-direction is twice the size of the semi-major axis of the original ellipse, measured as \(2a\).
• The axes in both the y and z directions are the same, stretching to \(2b\), matching twice the semi-minor axis of the ellipse.

Visualizing this rotational process helps in comprehending why the ellipsoid has these particular dimensions.
Imagining how the ellipse moves around the x-axis can shed light on how we gain the overall surface structure of the ellipsoid, with its rounded and elongated shape.
Understanding this revolving nature prepares us for diving deeper into mathematical calculations for surface area.

Integral Calculus

Integral Calculus comes into play when calculating the surface area of an ellipsoid. It allows us to sum up infinitesimally small pieces to get the total surface area.
The revolution method employs a specific formula to derive the surface area of a solid of revolution, which is given by:
\[ S = 2\pi \int_{0}^{a} y \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx \] This formula helps us compute the curved surface area generated by rotating the curve. Here, \(y\) represents the function of the ellipse we're dealing with.
The process often requires using techniques like substitution and recognizing patterns that align with standard integrals.
A proper change of variables, such as \(u=cx\), simplifies the integration, converting complex problems into more manageable ones.
These steps are essential for both understanding and computing the surface area efficiently.

Applications of Calculus

Applications of Calculus are vast, and they provide valuable insights into solving real-world problems. For the ellipsoid, calculus helps calculate its total surface area, which is vital in various fields, including engineering and physics.
From designing domes to determining volume capacities in vessels, understanding the exact surface measurements aids in better product design and safety.
Additionally, grasping these concepts promotes a better transition into more advanced calculus topics, where these fundamental ideas are applied to even more complex surfaces and forms.
The ellipsoidal surface area provides a concrete example of how calculus helps assess and solve challenges related to geometry and beyond.

• Geometers analyze surface variations using these calculated areas.
• Engineers use the data to optimize structures for both functionality and materials.

In essence, calculus transforms the theoretical into the practical, making it indispensable in professional fields.