# Solid of revolution

• Feb 17th 2009, 04:45 AM
Db0ne
Solid of revolution
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?
• Feb 17th 2009, 05:26 AM
Volume of revolution
Hello DbOne
Quote:

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 $\displaystyle x = 15$ between $\displaystyle y = 0$ and $\displaystyle y = 76$ is rotated about the $\displaystyle y$-axis.

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

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

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

• Feb 17th 2009, 05:41 AM
Db0ne
Quote:

Originally Posted by Grandad
Hello DbOneThe advice you have been given is quite correct.

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

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

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

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

so one of the equations would be $\displaystyle \pi \int \sqrt225-\sqrt(y-76)^2dy$? from 76 to 91
And the other one would be $\displaystyle \pi \int 15^2 dy$ from 0 to 76?
• Feb 17th 2009, 06:05 AM
Volume of revolution
Hello DbOne
Quote:

Originally Posted by Db0ne
so one of the equations would be $\displaystyle \pi \int \sqrt225-\sqrt(y-76)^2dy$? from 76 to 91
And the other one would be $\displaystyle \pi \int 15^2 dy$ from 0 to 76?

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

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

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

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

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

OK now?