Solid of revolution

**Db0ne** · February 17th 2009, 04:45 AM

I have to go around and find the volume of a silo-shaped trash can using solid of revolution
height 91cm
Circumference 119.3cm
Diameter 15cm
http://common.csnstores.com/United-R...R~UR1180_l.jpg is what the trash can looks like.

I need list of the equations(s) of the curve(s) which are going to be revolved around an axis. I also need to state the upper and lower limits for the equation(s). Then intergrate the curve(s)

I ask somebody else on line and they gave the equations intergral of x=15 from 0 to 76
and sqrt[225-(y-76)^2] from 76 to 91 is that right?

**Grandad** · February 17th 2009, 05:26 AM

Hello DbOne

Originally Posted by Db0ne

I have to go around and find the volume of a silo-shaped trash can using solid of revolution
height 91cm
Circumference 119.3cm
Diameter 15cm
http://common.csnstores.com/United-Receptacle-European-Designer-15-Gal.-Round-Top-Receptacle~img~UR~UR1180_l.jpg is what the trash can looks like.

I need list of the equations(s) of the curve(s) which are going to be revolved around an axis. I also need to state the upper and lower limits for the equation(s). Then intergrate the curve(s)

I ask somebody else on line and they gave the equations intergral of x=15 from 0 to 76
and sqrt[225-(y-76)^2] from 76 to 91 is that right?

The advice you have been given is quite correct.

The vertical sides are formed when the line $x = 15$ between $y = 0$ and $y = 76$ is rotated about the $y$ -axis.

The top is a hemisphere that is formed when a circle centre $(0, 76)$ radius $15$ , between $y = 76$ and $y = 91$ is rotated about the $y$ -axis. The equation of this circle is

$x^2 + (y-76)^2 = 15^2$

You'll need to find $\int \pi x^2 dy$ in each case, and add the two volumes together to get the total volume of the trash can.

Grandad

**Db0ne** · February 17th 2009, 05:41 AM

Originally Posted by Grandad

Hello DbOneThe advice you have been given is quite correct.

The vertical sides are formed when the line $x = 15$ between $y = 0$ and $y = 76$ is rotated about the $y$ -axis.

The top is a hemisphere that is formed when a circle centre $(0, 76)$ radius $15$ , between $y = 76$ and $y = 91$ is rotated about the $y$ -axis. The equation of this circle is

$x^2 + (y-76)^2 = 15^2$

You'll need to find $\int \pi x^2 dy$ in each case, and add the two volumes together to get the total volume of the trash can.

Grandad

so one of the equations would be $\pi \int \sqrt225-\sqrt(y-76)^2dy$ ? from 76 to 91
And the other one would be $\pi \int 15^2 dy$ from 0 to 76?

**Grandad** · February 17th 2009, 06:05 AM

Hello DbOne

Originally Posted by Db0ne

so one of the equations would be $\pi \int \sqrt225-\sqrt(y-76)^2dy$ ? from 76 to 91
And the other one would be $\pi \int 15^2 dy$ from 0 to 76?

Each integral is $\int \pi x^2 dy$ . So your second one is OK, because you've used $x = 15 \Rightarrow x^2 = 15^2$ .

But the first one comes from the equation of the circle:

$x^2 + (y-76)^2 = 15^2 =225$

i.e. $x^2 = 225 - (y-76)^2$ . So the integral is

$\int_{76}^{91}\pi (225 - (y-76)^2)dy$

OK now?

Grandad

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