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Chapter 6: Problem 6

Find the area of the surface generated when the given curve is revolved about the \(x\) -axis. $$y=12-3 x \text { on }[1,3]$$

Short Answer

Expert verified
Question: Find the surface area of the solid generated by revolving the curve \(y = 12 - 3x\) about the x-axis from x=1 to x=3. Answer: The surface area of the solid generated by revolving the curve is \(15\pi\sqrt{10}\).

Step by step solution

01

Identify the formula for surface area

Since we are revolving the curve about the x-axis, we will use the formula for surface area of revolution around the x-axis, which is given by: $$A = 2 \pi \int_{a}^{b} y\sqrt{1+(\frac{dy}{dx})^2} dx.$$
02

Differentiate the given function with respect to x

To find the derivative \(\frac{dy}{dx}\), differentiate the given function, \(y=12-3x\), with respect to x: $$\frac{dy}{dx} = -3$$
03

Find \(1+(\frac{dy}{dx})^2\)

Calculate \(1+(\frac{dy}{dx})^2\) to use in the integral: $$1+(\frac{dy}{dx})^2 = 1 + (-3)^2 = 1 + 9 = 10.$$
04

Set up the integral for the surface area

Now that we have the derivative and the value of \(1+(\frac{dy}{dx})^2\), we can set up the integral for the surface area using the formula mentioned in Step 1 and the given limits of integration: $$A = 2 \pi \int_{1}^{3} (12-3x)\sqrt{10} dx.$$
05

Evaluate the integral

Now, we will evaluate the integral to find the surface area: $$A = 2 \pi \int_{1}^{3} (12-3x)\sqrt{10} dx = 2 \pi (\sqrt{10}\int_{1}^{3} (12-3x) dx)$$ To evaluate the integral, first find the antiderivative of \((12-3x)\): $$F(x) = 12x - \frac{3}{2}x^2$$ Now, apply the limits of integration: $$F(3) - F(1) = [36 - \frac{9}{2} \cdot 9] - [12 - \frac{3}{2}] = -\frac{15}{2}$$ Now, substitute this value back into the surface area equation: $$A = 2 \pi (\sqrt{10} \cdot -\frac{15}{2})$$
06

Calculate the final value

Finally, calculate the surface area by evaluating the expression: $$A = 2 \pi (\sqrt{10} \cdot -\frac{15}{2}) = -15\pi\sqrt{10}$$ However, since the area cannot be negative, we must take the absolute value of the result: $$A = 15\pi\sqrt{10}$$

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