Question

1) The region enclosed by the x-axis, the line x=3 and the curve y = sq. Rt x is rotated about the x-axis. What is the volume of the solid generated? *

Answer 1

Using the disk method, the volume would be

$\displaystyle \pi \int_0^3 (\sqrt x)^2 \,\mathrm dx=\pi \int_0^3 x\,\mathrm dx=\frac\pi2 x^2\bigg|_0^3=\boxed{\frac{9\pi}2}$

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.

Answer 2

In this exercise we have to use the knowledge of the integral to calculate the volume of the disk that will correspond to:

$V= \frac{9\pi }{2}$

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:

$\pi \int\limits^3_0 {(\sqrt{x})^2} \, dx = \pi \int\limits^3_0 {x} \, dx = \frac{\pi x^2}{2} = \frac{9 \pi }{2}$

See more about volume at brainly.com/question/1578538