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Mathematics 15 Online
OpenStudy (anonymous):

A reservoir shaped like a right circular cone, point down, 20ft across the top and 8ft deep is filled to a depth of 5 feet. Water is to be pumped to the same level as the top. How much work does it take?

OpenStudy (owlfred):

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OpenStudy (annon):

density of water is 62.4 right?

OpenStudy (anonymous):

yes

OpenStudy (annon):

so you get your force

myininaya (myininaya):

can you tell me alittle about work? (i'm low on this knowledge) i know it equals force times distance but i don't see how to get force here?

OpenStudy (annon):

which will be 62.4 then you times it by 10/8y

myininaya (myininaya):

annon can you tell me how to get those two terms?

OpenStudy (annon):

ya the 62.4 is given or known, and you set the cone on a 2D graph

OpenStudy (annon):

the slope is y=8/10 make it x=10/8y since you need the y component

myininaya (myininaya):

oh ok density of water is always 62.4 ok got it

OpenStudy (annon):

multiply them

myininaya (myininaya):

you're so smart i don't understand work at all it must be a physics/engineer thing

OpenStudy (annon):

i am a mech engineering major :)

myininaya (myininaya):

i was math major

OpenStudy (annon):

work = (6240/64)pi integral from 0 to 10 of y^3 dy

OpenStudy (annon):

did u get it mdntjem?

OpenStudy (anonymous):

I have a question, how did you get the limits of integration from 0 to 10?

OpenStudy (annon):

well you have to pump from bottom to top right?

OpenStudy (anonymous):

yes

OpenStudy (annon):

thats how limits of integration is where it begins to where it en

OpenStudy (anonymous):

I thought it would be to 8

OpenStudy (annon):

it would 8 if they said to stop at 8

OpenStudy (annon):

thats how limits of integration is where it begins to where it en

OpenStudy (anonymous):

Is it because the cone is parallel to the y-axis and the upper limit needs to be the top line perpendicular to the y-axis?

OpenStudy (annon):

if you look at the cone from a side veiw

OpenStudy (annon):

oh wait, ur right 0 to 8

OpenStudy (annon):

sorry lol and my numbers are wrong give me a sec.

OpenStudy (anonymous):

Okay :) no prob

OpenStudy (annon):

sorry about that :)

OpenStudy (annon):

62.4(400/64)*pi integral from 0 to 8 of y^3 dy

OpenStudy (anonymous):

Not a problem

OpenStudy (annon):

this should be the right answer after you integrate it out

OpenStudy (anonymous):

I have another quick question, how'd you get the (400/64)?

OpenStudy (annon):

if you look at the cone from the side, it makes a right triangle on a graph and it starts at zero and slopes up 8 and over 20

OpenStudy (annon):

so your equation is y=(8/20) x

OpenStudy (annon):

since you need it in terms of x to be your radius, you solve for x, x= (20/8)y

OpenStudy (annon):

now you use this as your "r" for the formula of the volume.

OpenStudy (annon):

pi*r^2

OpenStudy (annon):

so... pi* ((20/8)y)^2

OpenStudy (annon):

so you get pi* (20^2 / 8^2) y^2

OpenStudy (anonymous):

Oh I see!

OpenStudy (annon):

:D i hope i was clear enough

OpenStudy (anonymous):

I was doing mine with 10/8, so I got 5/4x=r

OpenStudy (annon):

oh i did that too at first until i realised it was a right conical tank

OpenStudy (anonymous):

Okay, I get it now. Thank you for all your help!

OpenStudy (annon):

np :)

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