# How do you find the volume of the solid generated by revolving #y=10/x^2#, #y=0#, #x=1#, #x=5# about the y-axis?

I hope it is not too confusing:

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To find the volume of the solid generated by revolving the region bounded by the curves (y = \frac{10}{x^2}), (y = 0), (x = 1), and (x = 5) about the y-axis, you can use the disk method.

The formula for the volume of revolution using the disk method is:

[V = \pi \int_{a}^{b} [f(x)]^2 dx]

Where (f(x)) represents the function generating the shape, and ([f(x)]^2) represents the area of a circular disk with radius (f(x)).

First, find the points of intersection of the curves (y = \frac{10}{x^2}) and (y = 0). This occurs when (\frac{10}{x^2} = 0), which implies (x) can't be 0. Thus, the intersection points are when (x = 1) and (x = 5).

Now, integrate from (x = 1) to (x = 5):

[V = \pi \int_{1}^{5} \left(\frac{10}{x^2}\right)^2 dx]

[= \pi \int_{1}^{5} \frac{100}{x^4} dx]

[= \pi \left[-\frac{100}{3x^3}\right]_{1}^{5}]

[= \pi \left(-\frac{100}{3(5^3)} + \frac{100}{3(1^3)}\right)]

[= \pi \left(-\frac{100}{375} + \frac{100}{3}\right)]

[= \pi \left(-\frac{4}{15} + \frac{100}{3}\right)]

[= \pi \left(\frac{100}{3} - \frac{4}{15}\right)]

[= \pi \left(\frac{500}{15} - \frac{4}{15}\right)]

[= \pi \left(\frac{496}{15}\right)]

[= \frac{496\pi}{15}]

So, the volume of the solid generated by revolving the region about the y-axis is (\frac{496\pi}{15}) cubic units.

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