Calc II question:
Given is a picture of a shed that is a half cylinder with radius as "r" and length as "l."
What is the volume of the shed? I found to be (pi*r^2*l)/2
The shed is ﬁlled with sawdust whose density (mass/unit volume) at any point is proportional to the distance of that point from the ﬂoor. The constant of proportionality is k. Calculate the total mass of sawdust in the shed.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- IrishBoy123

guessing without a drawing, think you're looking at \(M = \iiint \rho \ dV = \iiint \ (kr sin \theta) \ r \ dr \ \ d \theta \ dz\)
http://www.wolframalpha.com/input/?i=%5Cint_%7Bz%3D0%7D%5E%7BL%7D+%5Cint_%7B%5Ctheta+%3D+0%7D%5E%7B%5Cpi%7D+%5Cint_%7Br+%3D+0%7D%5E%7BR%7D+k+r+sin+%5Ctheta+r+dr+d%5Ctheta+dz

- IrishBoy123

i am travelling today so might be of limited use to you if you are pushing ahead; but there should be loads of clues in there, especially in the integral formed in wolfram; and @Phi is "the" go-to for this kind of stuff.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.