1. ## 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?

2. ## Volume of revolution

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.

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.

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?

4. ## Volume of revolution

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?