# Thread: Stuck on Volume of Solids

1. ## Stuck on Volume of Solids

Hey guys, I have been attempting the following problem for days now, and I just cannot seem to get the right answers. I know it probably is really simple and that I am over thinking it, but my answers are always off. I would really appreciate the help. Thanks!

Find the volume of the solid generated by resolving the region bounded by the following information:

y=sqrt(x), y=0, x=4

a) the x-axis
b) the y-axis
c) the line x=4
d) the line x=6

2. You mean revolving the region around each of those 4 lines.

Unfortunately, there are many different ways to do that and you don't say which way you are practicing. I would recommend the "disk" method for (a) and (c) and the "washer" method for (b) and (d). What do you know about those methods?

In (a), where the axis is horizontal you will be integrating with respect to x. In the others the axis is vertical and you will be integrating with respect to y.

3. Originally Posted by thatdancerale
Hey guys, I have been attempting the following problem for days now, and I just cannot seem to get the right answers. I know it probably is really simple and that I am over thinking it, but my answers are always off. I would really appreciate the help. Thanks!

Find the volume of the solid generated by resolving the region bounded by the following information:

y=sqrt(x), y=0, x=4

a) the x-axis
b) the y-axis
c) the line x=4
d) the line x=6
In general to get the volume for these types of problems, you integrate the cross-sectional area from the bounds a to b.

$V = \int_a^b A(x)dx$
This is called the method of disks.

a) The region they're referring to is the area underneath y = sqrt(x) from 0 to 4. If you revolve it around the x-axis, you'll get some sort of a cone-shaped figure (use your imagination), with a circular cross-section. The area of the cross-section is

$A = \pi r^2$ << the area of a circle.

In this case, the radius, r, is f(x) so the area is
$A = \pi [f(x)]^2$

therefore the volume will be
$V = \int_0^4 \pi [\sqrt{x}]^2dx$
$V = \pi\int_0^4xdx$
$V = \pi\left[\frac{x^2}{2}\right]_0^4$
$V = 8\pi$

b) In (a) we used the method of disks. Here, we're going to the method of cylindrical shells. Read about it to get an idea of how it works

$V = \int_0^4 2\pi x\sqrt{x}dx$
$V = 2\pi\int_0^4x^\frac{3}{2}dx$
$V = 2\pi\left[\frac{2x^\frac{5}{2}}{5}\right]_0^4$
$V = \frac{128\pi}{5}$

c) Here, it's a special case because we're not revolving it around the central axes. So instead imagine x = 4 as being the y-axis. So using the method of cylindrical shells
$V = 2\pi\int_0^4(4-x)(\sqrt{x})dx$
$V = 2\pi\int_0^44x^\frac{1}{2} - x^\frac{3}{2}dx$
$V = 2\pi\left[\frac{8x^\frac{3}{2}}{3} - \frac{2x^\frac{5}{2}}{5}\right]_0^4$
$V = \frac{256\pi}{15}$

The reason i changed the (x) in the (b) to (4 - x) here, is because the x-distance from x = 4 to the region under y = sqrt(x) is no longer just the (x) value; it's now (4-x). I didn't change the sqrt(x) to anything, because the height still remains sqrt(x). If you don't understand this, read about the method of cylindrical shells.

d) Do the last one. It's verrry similar to (c). Show me your answer. (hint: read the what i said just before this in (c))

Anyone correct me if i'm wrong, thanks