Finding the volume with Shell method

[denian]

the volume that i get is a negative value, there must be something wrong with my working.
is shell height = y + 2 - y^2
or i am wrong?
tq.

 

[Fermat]

The shell Method:
When taking a volume of revolution about the y-axis, the formula is,

$$V = \int_{x_0} ^{x_1} xy \ dx$$

where x is the radius of the shell, dx represents the shell thickness and y is the length, or height, of the shell.

Since, in this case, you are taking negative y-values, then you will get a negative answer for the volume. Simply change the sign.

You have two curves to contend with, so you should make up the integral like this,

$$V = \int_0 ^1 xy_1 \ dx + \int_1 ^2 xy_2 \ dx$$

$$\mbox{where}\ y_1\ \mbox{is}\ \sqrt{x}\ \mbox{and}\ y_2\ \mbox{is}\ (x-2).$$

 

[denian]

thank you, but I am a bit confused.
im doing some self-study here, and the formula they wrote in the book is what i wrote in the first line.
btw, the x-axis is the axis of rotation.

nvm. i try to figure it out again :)

 

[amcavoy]

The shell Method:
When taking a volume of revolution about the y-axis, the formula is,

$$V = \int_{x_0} ^{x_1} xy \ dx$$

where x is the radius of the shell, dx represents the shell thickness and y is the length, or height, of the shell.

Since, in this case, you are taking negative y-values, then you will get a negative answer for the volume. Simply change the sign.

You have two curves to contend with, so you should make up the integral like this,

$$V = \int_0 ^1 xy_1 \ dx + \int_1 ^2 xy_2 \ dx$$

$$\mbox{where}\ y_1\ \mbox{is}\ \sqrt{x}\ \mbox{and}\ y_2\ \mbox{is}\ (x-2).$$

...don't you mean:

$$V=\mathbf{2\pi}\int_{a}^{b}xf(x)dx$$

 

[Fermat]

...don't you mean:

$$V=\mathbf{2\pi}\int_{a}^{b}xf(x)dx$$

yes,

thanks for the correction,

y = f(x)

and I completely missed out the 2pi !

 

[Fermat]

thank you, but I am a bit confused.
im doing some self-study here, and the formula they wrote in the book is what i wrote in the first line.
btw, the x-axis is the axis of rotation.

nvm. i try to figure it out again :)

Sorry about the confusion there. I'm afraid I misread your work. Actually, I'd never heard of the shell method before.

You working is correct. The shell height/length is (y + 2 - y²). And you will have gotten a volume of -5pi/6 units, yes?
I got a volume of +5pi/6 units using my volume of revolution method - which I've now found out is called the disk method or washer method.

You are getting a negative value simply because you are using negative y-values for the radius, but +ve values for the shell height.