Volume Integration Around Non-Coordinate Axis

Homework Statement

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x = 5. (Use disk method)

$$xy = 3, y = 1, y = 4, x = 5$$

Homework Equations

[/B]
The formula using for disk method is of the form:

$$\pi \int (r(x/y))^2*(dx/y)$$

The Attempt at a Solution

[/B]
So first I graphed this out, and came up with a region, like so (represented by the white region):

In disk method, when rotating around a vertical axis, the differential of dy is used.

Setting this integral up, our limits are y = 1, to y = 4, as given in the problem statement.

My main question is how do we represent the r in the disk method equation, when the axis is meant to be around x = 5, and not the y-axis.

The best I could think of would be:

$$\pi \int_{1}^{4} (\frac {y} {3} + 5)^{2} dy$$

However, something tells me this isn't the right approach, because I'm not seeing the logic behind whether or not I should add, subtract, or subtract from the x = 5.

Thank you.

SteamKing
Staff Emeritus
Science Advisor
Homework Helper

Homework Statement

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x = 5. (Use disk method)

$$xy = 3, y = 1, y = 4, x = 5$$

Homework Equations

[/B]
The formula using for disk method is of the form:

$$\pi \int (r(x/y))^2*(dx/y)$$

The Attempt at a Solution

[/B]
So first I graphed this out, and came up with a region, like so (represented by the white region):

In disk method, when rotating around a vertical axis, the differential of dy is used.

Setting this integral up, our limits are y = 1, to y = 4, as given in the problem statement.

My main question is how do we represent the r in the disk method equation, when the axis is meant to be around x = 5, and not the y-axis.

The best I could think of would be:

$$\pi \int_{1}^{4} (\frac {y} {3} + 5)^{2} dy$$

However, something tells me this isn't the right approach, because I'm not seeing the logic behind whether or not I should add, subtract, or subtract from the x = 5.

Thank you.
Remember, r(y) is the distance measured from the curve xy = 3 to the line x = 5. Examining your graph should tell you what r(1) must be. I don't think that's what the integrand in your integral indicates.

SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member

Homework Equations

[/B]
The formula using for disk method is of the form:$$\pi \int (r(x/y))^2*(dx/y)$$
I suspect that the formula you gave for the disk method is to be interpreted as a general formula in which you will use only x or only y , depending upon whether the axis is horizontal or is vertical .

Use ##\displaystyle \ \ \pi \int (r(x))^2\ dx\ \ ## or ##\displaystyle \ \ \pi \int (r(y))^2\ dy\ \ ## as appropriate.

So, you could interpret the distance as ## 5 - x ##? and thus make the ## r = 5 - \frac 3 y ## ?

The integral would then become ## \pi \int_{1}^{4} (5 - \frac {3} {y} )^{2} dy ## ?

SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
So, you could interpret the distance as ## 5 - x ##? and thus make the ## r = 5 - \frac 3 y ## ?

The integral would then become ## \pi \int_{1}^{4} (5 - \frac {3} {y} )^{2} dy ## ?
Yes. Looks good to me.