# What is the surface area produced by rotating #f(x)=x^2lnx, x in [0,3]# around the x-axis?

If we consider a small strip width

The arc length

With

and so the surface area of the element is

For

However because

The surface area in total is therefore

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To find the surface area produced by rotating ( f(x) = x^2 \ln(x) ) around the x-axis over the interval [0, 3], you can use the formula for the surface area of a curve rotated about the x-axis:

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

Where ( \frac{dy}{dx} ) is the derivative of ( f(x) ) with respect to ( x ).

First, find ( \frac{dy}{dx} ): [ f(x) = x^2 \ln(x) ] [ \frac{dy}{dx} = \frac{d}{dx}(x^2 \ln(x)) ]

Using the product rule: [ \frac{dy}{dx} = 2x \ln(x) + x ]

Now, plug ( f(x) ) and ( \frac{dy}{dx} ) into the surface area formula and integrate over the interval [0, 3]:

[ S = 2\pi \int_{0}^{3} x^2 \ln(x) \sqrt{1 + (2x \ln(x) + x)^2} ,dx ]

After evaluating this integral, you will find the surface area produced by rotating ( f(x) = x^2 \ln(x) ) around the x-axis over the interval [0, 3].

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