how do I set up the integral to find the volume of the solid obtained by revolving the region about the x-axis when y=square root of x, y=1 and x=4

Question

anonymous · Answer

I think it turns into a disk but I am not sure. And I cant figure out the integral.

dumbcow · Answer

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this uses the washer method where for each circular cross-section you take large area and subtract smaller area
Here the outer radius is sqrt(x), the inner radius is 1
$$V = \pi \int\limits_{?}^{?}R^{2} -r^{2} $$
$$V = \pi \int\limits_{1}^{4}(\sqrt{x}^{2} - 1^{2}) dx =  \pi \int\limits_{1}^{4}(x - 1)dx$$