# What is the surface area of the solid created by revolving #f(x) =ln(4-x) , x in [1,3]# around the x axis?

Surface area of the solid created by revolving

graph{ln(4-x) [-0.99, 4.01, -0.52, 1.98]}

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To find the surface area of the solid created by revolving the function ( f(x) = \ln(4 - x) ) around the x-axis over the interval ([1, 3]), you'll use the formula for the surface area of a solid of revolution.

The formula for the surface area of a solid of revolution is:

[ A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left( \frac{dy}{dx} \right)^2} , dx ]

Given ( f(x) = \ln(4 - x) ), the derivative with respect to ( x ) is:

[ \frac{dy}{dx} = -\frac{1}{4 - x} ]

Substitute these into the formula and integrate over the interval ([1, 3]):

[ A = 2\pi \int_{1}^{3} \ln(4 - x) \sqrt{1 + \left( -\frac{1}{4 - x} \right)^2} , dx ]

Now, evaluate the integral. This involves simplifying the expression under the square root and then integrating the resulting expression.

[ A = 2\pi \int_{1}^{3} \ln(4 - x) \sqrt{1 + \frac{1}{(4 - x)^2}} , dx ]

[ A = 2\pi \int_{1}^{3} \ln(4 - x) \sqrt{\frac{(4 - x)^2 + 1}{(4 - x)^2}} , dx ]

[ A = 2\pi \int_{1}^{3} \ln(4 - x) \sqrt{\frac{(16 - 8x + x^2 + 1)}{(4 - x)^2}} , dx ]

[ A = 2\pi \int_{1}^{3} \ln(4 - x) \sqrt{\frac{(x^2 - 8x + 17)}{(4 - x)^2}} , dx ]

[ A = 2\pi \int_{1}^{3} \ln(4 - x) \frac{\sqrt{x^2 - 8x + 17}}{|4 - x|} , dx ]

This integral might require numerical methods to evaluate. Once evaluated, it will give you the surface area of the solid of revolution.

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