Using the disk method, the volume would be
[tex]\displaystyle \pi \int_0^3 (\sqrt x)^2 \,\mathrm dx=\pi \int_0^3 x\,\mathrm dx=\frac\pi2 x^2\bigg|_0^3=\boxed{\frac{9\pi}2}[/tex]
That is, you split up the solid into disks of some small height ∆x, and each disk has radius at point x equal to the distance (√(x)) from the axis of revolution (the x-axis, y = 0) to the curve y = √(x). The volume of such a disk is then π (√(x))² ∆x. Take the sum of these volumes to get the total volume, then as the heights get smaller and smaller, replace ∆x with dx and the sum with the integral.
In this exercise we have to use the knowledge of the integral to calculate the volume of the disk that will correspond to:
[tex]V= \frac{9\pi }{2}[/tex]
That is, you split up the solid into disks of some small height ∆x, and each disk has radius at point x equal to the distance (√(x)) from the axis of revolution (the x-axis, y = 0) to the curve y = √(x). The volume of such a disk is then π (√(x))² ∆x. Take the sum of these volumes to get the total volume, then as the heights get smaller and smaller, replace ∆x with dx and the sum with the integral. Using the disk method, the volume would be:
[tex]\pi \int\limits^3_0 {(\sqrt{x})^2} \, dx = \pi \int\limits^3_0 {x} \, dx = \frac{\pi x^2}{2} = \frac{9 \pi }{2}[/tex]
See more about volume at brainly.com/question/1578538
