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- Thread starter denian
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Fermat

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When taking a volume of revolution about the y-axis, the formula is,

[tex]V = \int_{x_0} ^{x_1} xy \ dx[/tex]

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,

[tex]V = \int_0 ^1 xy_1 \ dx + \int_1 ^2 xy_2 \ dx[/tex]

[tex]\mbox{where}\ y_1\ \mbox{is}\ \sqrt{x}\ \mbox{and}\ y_2\ \mbox{is}\ (x-2).[/tex]

- #3

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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 :)

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Fermat said:

When taking a volume of revolution about the y-axis, the formula is,

[tex]V = \int_{x_0} ^{x_1} xy \ dx[/tex]

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,

[tex]V = \int_0 ^1 xy_1 \ dx + \int_1 ^2 xy_2 \ dx[/tex]

[tex]\mbox{where}\ y_1\ \mbox{is}\ \sqrt{x}\ \mbox{and}\ y_2\ \mbox{is}\ (x-2).[/tex]

...don't you mean:

[tex]V=\mathbf{2\pi}\int_{a}^{b}xf(x)dx[/tex]

- #5

Fermat

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yes,apmcavoy said:...don't you mean:

[tex]V=\mathbf{2\pi}\int_{a}^{b}xf(x)dx[/tex]

thanks for the correction,

y = f(x)

and I completely missed out the 2pi !!

- #6

Fermat

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

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 :)

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

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

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