To find the exact area of the surface obtained by rotating the curve \(y = 1 + 3x\) about the x-axis from \(x = 1\) to \(x = 3\), we use the formula for the surface area of revolution about the x-axis:
\[ S = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx \]
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Find \(\frac{dy}{dx}\): The derivative of \(y\) with respect to \(x\) is: \[ \frac{dy}{dx} = 3 \]
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Set up the integral: Now we need to compute \(\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\): \[ \left(\frac{dy}{dx}\right)^2 = 3^2 = 9 \] Therefore: \[ \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{1 + 9} = \sqrt{10} \]
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Plug everything into the surface area formula: Substituting \(y = 1 + 3x\) and \(\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{10}\) into the integral, we get: \[ S = 2\pi \int_1^3 (1 + 3x) \sqrt{10} , dx \] This can be simplified to: \[ S = 2\pi \sqrt{10} \int_1^3 (1 + 3x) , dx \]
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Evaluate the integral: We can split the integral into two parts: \[ \int_1^3 (1 + 3x) , dx = \int_1^3 1 , dx + \int_1^3 3x , dx \]
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The first integral: \[ \int_1^3 1 , dx = [x]_1^3 = 3 - 1 = 2 \]
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The second integral: \[ \int_1^3 3x , dx = 3 \left[\frac{x^2}{2}\right]_1^3 = 3 \left[\frac{9}{2} - \frac{1}{2}\right] = 3 \cdot \frac{8}{2} = 12 \]
So, we have: \[ \int_1^3 (1 + 3x) , dx = 2 + 12 = 14 \]
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Substituting back into the surface area formula: \[ S = 2\pi \sqrt{10} (14) = 28\pi \sqrt{10} \]
Thus, the exact area of the surface obtained by rotating the curve about the x-axis is:
\[ \boxed{28\pi \sqrt{10}} \]
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