Volume of revolution integral question

[arc1]

Homework Statement

This project deals with custom made gold wedding bands. Its shape is obtained by revolving the region shown about a horizontal axis. The resulting band has Inner radius R, Minimum Thickness T, Width W.

The curved boundary of the region is an arc of a circle whose center lies on the axis of revolution. For a typical wedding band with given dimensions R, T, W, the jeweller must first calculate the volume of the desired wedding band to determine how much gold will be required.

Show that the volume V is given by the formula

$$\frac {\pi W}{6} (W^2 + 12RT + 6T^2)$$

Homework Equations

$$x^2 + y^2 = R^2$$

The Attempt at a Solution

First i tried to use the shell method and came up with this:

$$2 \cdot 2\pi \left(\int_{R}^{(R+T)}x\frac w2 \,dx + \int_{(R+T)}^{b}x\sqrt{b^2-x^2} \, dx \right)$$

where b is the radius of the curved boundary

and that led to this:

$$\pi W(2RT+T^2) + \pi\frac 43(b^2 -(R+T)^2)$$

But now i got stuck. I don't know what to do about the b.
I don't even know if I'm on the right track now.

 

[HallsofIvy]

Homework Statement

This project deals with custom made gold wedding bands. Its shape is obtained by revolving the region shown about a horizontal axis. The resulting band has Inner radius R, Minimum Thickness T, Width W.

The curved boundary of the region is an arc of a circle whose center lies on the axis of revolution. For a typical wedding band with given dimensions R, T, W, the jeweller must first calculate the volume of the desired wedding band to determine how much gold will be required.

Show that the volume V is given by the formula

$$\frac {\pi W}{6} (W^2 + 12RT + 6T^2)$$

Homework Equations

$$x^2 + y^2 = R^2$$

This is confusing. In your picture "R" appears to be distance from the axis of revolution to the inner edge of the "ring". If that is true, then R is not the radius of the circle forming the outer edge. If R is the radius of that circle (as in the equation just above) what is the distance from the inner edge to the axis of revolution?

The Attempt at a Solution

First i tried to use the shell method and came up with this:

$$2 \cdot 2\pi \left(\int_{R}^{(R+T)}x\frac w2 \,dx + \int_{(R+T)}^{b}x\sqrt{b^2-x^2} \, dx \right)$$

where b is the radius of the curved boundary

and that led to this:

$$\pi W(2RT+T^2) + \pi\frac 43(b^2 -(R+T)^2)$$

But now i got stuck. I don't know what to do about the b.
I don't even know if I'm on the right track now.

 

[Dick]

Good start, you've got the integral set up right. To find b draw a right triangle whose hypotenuse connects the center point on the axis to the endpoint of the arc. One side is (R+T), the other side is W/2 and the hypotenuse is b. Now you can find b. And did you mean to have a 3/2 power on the b^2-(R+T)^2?

 

[Dick]

This is confusing. In your picture "R" appears to be distance from the axis of revolution to the inner edge of the "ring". If that is true, then R is not the radius of the circle forming the outer edge. If R is the radius of that circle (as in the equation just above) what is the distance from the inner edge to the axis of revolution?

If you look at the working of the problem the OP is using b to denote the radius of the arc. Probably should have labeled that in the picture. The problem is actually almost done!

 

[arc1]

Thanks Dick.
Your help made it easy.

And you're right it was supposed to be like this:

$$V = \pi W(2RT+T^2) + \pi\frac 43(b^2 -(R+T)^2)^{\frac 32}$$

And I continue:

$$b^2 = (R+T)^2 + (\frac W2)^2$$
$$V = \pi W(2RT+T^2) + \pi\frac 43((R+T)^2 + \left(\frac W2\right)^2 -(R+T)^2)^{\frac 32}$$

$$V = \pi W(2RT+T^2) + \pi\frac 43\left(\frac W2\right)^{2 \cdot \frac 32}$$

$$V = \pi \left(W(2RT+T^2) + \frac {W^3}{6}\right)$$

$$\underline{\underline{V = \frac{\pi W}{6} \left(W^2 + 12RT + 6T^2 \right)}}$$

 

[Dick]

Bingo. You are pretty good.