Problem 2 - Calculus Early Transcendentals

Question

A frustum of a cone is generated by revolving the graph of \(y=4 x\) on the interval [2,6] about the \(x\) -axis. What is the area of the surface of the frustum?

Step-by-Step Solution

Answer

Question: Find the total surface area of the frustum generated by revolving the graph of \(y=4x\) on the interval \([2,6]\) about the x-axis. Solution: The total surface area of the frustum is given by: $$A_{total} = (136\pi \sqrt{17} + 576\pi + 64\pi)$$

Step 1: Find the function describing the lateral surface area

To find the lateral surface area, we can use the formula: $$A_{lateral} = 2\pi \int_a^b y \sqrt{1 + (\frac{dy}{dx})^2} dx$$ Where \(y\) is the function and \(dy/dx\) is its derivative. In our case, \(y=4x\) and the interval is \([2,6]\). First, let's find the derivative of \(y\) with respect to \(x\): $$\frac{dy}{dx} = \frac{d(4x)}{dx} = 4$$ Now we will find \((\frac{dy}{dx})^2\): $$\left(\frac{dy}{dx}\right)^2 = (4)^2 = 16$$ So, our expression for the surface area becomes: $$A_{lateral} = 2\pi \int_2^6 4x \sqrt{1 + 16} dx$$

Step 2: Solve the integral

Now we have to solve the integral: $$A_{lateral} = 2\pi \int_2^6 4x \sqrt{17} dx$$ Since \(\sqrt{17}\) is a constant, we can pull it out of the integral: $$A_{lateral} = 2\pi \sqrt{17} \int_2^6 4x dx$$ Now we need to find the antiderivative of \(4x\): $$\int 4x dx = 2x^2 + C$$ Now, apply the limits of integration: $$A_{lateral} = 2\pi \sqrt{17}( (2\cdot 6^2) - (2\cdot 2^2) ) = 2\pi \sqrt{17}( 68 )$$ So, the lateral surface area is: $$A_{lateral} = 136\pi \sqrt{17}$$

Step 3: Calculate the area of top and bottom circular surfaces

Since the frustum is generated by revolving the graph \(y=4x\) about the x-axis, the radius of the top surface is the value of \(y\) when \(x=6\), and the radius of the bottom surface is the value of \(y\) when \(x=2\). Top surface radius: \(y(6) = 4\cdot 6 = 24\) Bottom surface radius: \(y(2) = 4\cdot 2 = 8\) Now we will calculate the areas of the top and bottom circular surfaces using the formula for the area of a circle, \(A=\pi r^2\). Top surface area: \(A_{top} = \pi (24^2) = 576\pi\) Bottom surface area: \(A_{bottom} = \pi (8^2) = 64\pi\)

Step 4: Find the total surface area

Now, we can find the total surface area of the frustum by adding together the lateral surface area, top surface area, and bottom surface area: Total surface area: $$A_{total} = A_{lateral} + A_{top} + A_{bottom} = 136\pi \sqrt{17} + 576\pi + 64\pi$$ Thus, the total surface area of the frustum is: $$A_{total} = (136\pi \sqrt{17} + 576\pi + 64\pi)$$