Problem 35 - Thomas Calculus

Question

A cone frustum The line segment joining the points \((0,1)\) and \((2,2)\) is revolved about the \(x\) -axis to generate a frustum of a cone. Find the surface area of the frustum using the parametrization \(x=2 t, y=t+1,0 \leq t \leq 1 .\) Check your result with the geometry formula: Area \(=\pi\left(r_{1}+r_{2}\right)(\) slant height\().\)

Step-by-Step Solution

Answer

The surface area is \(3\pi\sqrt{5}\).

Step 1: Find the Radii

When the line segment is revolved around the x-axis, it forms a frustum of a cone. The distances from the x-axis to the points on the line segment become the radii. For \(t=0\), \(y=0+1=1\), so \(r_1=1\). For \(t=1\), \(y=1+1=2\), so \(r_2=2\).

Step 2: Determine the Slant Height

To find the slant height \(l\) of the frustum, calculate the distance between the endpoints of the line segment in the 2D plane before it is revolved. Using the distance formula \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\), we have \(d=\sqrt{(2-0)^2+(2-1)^2}=\sqrt{4+1}=\sqrt{5}\). So, the slant height \(l=\sqrt{5}\).

Step 3: Apply the Surface Area Formula for a Frustum

The surface area \(A\) of the frustum is given by \(A=\pi (r_1+r_2)(\text{slant height})\), where \(r_1\) and \(r_2\) are the radii found earlier, and the slant height is as calculated. Substituting we get: \(A=\pi (1+2)\sqrt{5}=3\pi\sqrt{5}\).

Step 4: Verify with Parametrization

Using the surface area formula from revolved curves, \(A= \int_{a}^{b} 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\). For this parametrization: \(x=2t\) and \(y=t+1\). Thus \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=1\). Substitute into the integral:\[ A = \int_{0}^{1} 2\pi (t+1) \sqrt{2^2 + 1^2} \, dt = 2\pi \sqrt{5} \int_{0}^{1} (t+1) \, dt. \]Evaluate the integral: \[ \int_{0}^{1} (t+1) \, dt = \left[ \frac{t^2}{2} + t \right]^1_0 = \frac{1^2}{2} + 1 - \left(\frac{0^2}{2} + 0\right) = \frac{3}{2}. \]Finally, calculate \(A = 2\pi \sqrt{5} \cdot \frac{3}{2} = 3\pi \sqrt{5}\). This matches the geometric formula result.

Key Concepts

Parametrization

Let's explore how we can use parametrization to model geometric shapes such as a frustum of a cone. Parametrization is the process of defining a set of equations to describe a geometric object. In this exercise, we are given the parametrization equations: \(x = 2t\) and \(y = t + 1\), where \(0 \leq t \leq 1\).

These equations tell us how to map a parameter \(t\) onto points \((x, y)\) on a planar line segment. By revolving this segment around the \(x\)-axis, we create a 3D shape - a frustum of a cone.

• The parameter \(t\) varies from 0 to 1, tracing the line segment entirely.
• For \(t=0\), the point \((x, y)\) is \((0, 1)\).
• For \(t=1\), the point \((x, y)\) is \((2, 2)\).

As \(t\) changes, these equations determine the values of \(x\) and \(y\), defining the curve that, when revolved, forms the frustum. Parametrization simplifies complex shapes into manageable mathematical expressions, letting us apply various calculus concepts effectively.

Surface Area Calculation

Calculating the surface area of a frustum involves both geometric intuition and calculus. Here, the surface area \(A\) is expressed through the formula involving its radii and slant height: \[ A = \pi (r_1 + r_2) \times \text{slant height} \].

• First, we determine the radii by examining the parametrization at \(t=0\) and \(t=1\): \(r_1 = 1\) and \(r_2 = 2\).
• Next, we calculate the slant height as the distance between the endpoints \((0, 1)\) and \((2, 2)\): \(\text{slant height} = \sqrt{5}\).

Additionally, we verify the surface area using the calculus-based approach for finding the area of a surface of revolution:
\[ A = \int_{0}^{1} 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \].
Substituting \( y = t + 1\), \( \frac{dx}{dt} = 2\), and \( \frac{dy}{dt} = 1\), we evaluate the integral to get the correct surface area. It's amazing to see how calculus seamlessly connects geometry, confirming our surface area finding \( 3\pi \sqrt{5} \).

Geometry Formula Verification

After calculating the surface area, it's useful to verify our result using known geometry formulas. The geometric formula for a frustum's area is \( A = \pi (r_1 + r_2)(\text{slant height}) \). Verifying involves substituting the calculated values into this formula:

• Radii: \(r_1 = 1\) and \(r_2 = 2\).
• Slant height: \(\sqrt{5}\).

Putting these values into the formula yields:
\[ A = \pi (1 + 2) \cdot \sqrt{5} = 3\pi \sqrt{5} \].

We achieved the same result using parametrization and the calculus approach. This consistency shows how both geometric intuition and calculus methods can lead us to the same conclusion. It demonstrates the power of mathematical verification, ensuring our understanding of frustum surface area calculations is robust and reliable. In this case, both methods corroborate each other, confirming the accuracy of the calculated surface area.