Integral Calculus: Revolutions

[dmanz]

I'm having trouble with these revolutions questions. (see attached)

any help would be greatly appreciated. i will love you long time

(made with Paint : Don't hate me)

[dmanz]

bump. oh okay i see how it is.
 

.....only kidding i really need help. like real bad.

[xZero]

umm...area of what? and how is that a bowl?

[dmanz]

umm...area of what? and how is that a bowl?

i can't express A in terms of h and r

umm the minimum surface area of the cone when it is rotated along the x axis.
it is like a cone-shaped bowl.

[xZero]

okay to make it clear to myself and everyone else, type out what you want us to find. The diagrams fine, its just that I have no fkn idea what the question is asking me to find

[pi]

The equation you have given looks more like a surface area of revolution rather than a volume of revolution...

EDIT: What is 'y'? Do we need an equation?

[dmanz]

okay to make it clear to myself and everyone else, type out what you want us to find. The diagrams fine, its just that I have no fkn idea what the question is asking me to find

sorry man, ill try and clear it up

Surface area of the cone= 2000cm^3

1. Express A  in terms of h and r,  where y= r/h x (use the eqn A= integral...h, 0 in diagram)
2. show that A=..... (second eqn)
3. Find dimensions of the cone (in mm)
4. what is  the minimum area of the cone (in cm^2)

[dmanz]

The equation you have given looks more like a surface area of revolution rather than a volume of revolution...

EDIT: What is 'y'? Do we need an equation?

when y = r/h x

[xZero]

so you want us to solve it from q3 or start from stratch?

and also for q3, you want us to find a specific dimension of the cone or just the range of values for h and r?

[dmanz]

so you want us to solve it from q3 or start from stratch?

and also for q3, you want us to find a specific dimension of the cone or just the range of values for h and r?

but i haven't done 1 or 2 properly yet. like when i sub the y and dy/dx in the eqn i don't get the same answer on the calc.
for question 3 the question asks for h, r and s

sorry bout my explainations man i suck with computers

[xZero]

q1. http://www.wolframalpha.com/input/?i=integrate+2*pi*r*x*sqrt%281%2Br^2%2Fh^2%29%2Fh+dx+from+0+to+h

q2. pretty sure theres an error in your picture, i got

[dmanz]

q1. http://www.wolframalpha.com/input/?i=integrate+2*pi*r*x*sqrt%281%2Br^2%2Fh^2%29%2Fh+dx+from+0+to+h

q2. pretty sure theres an error in your picture, i got

cheers for q1.
how did you get your ans in q2?

[xZero]

for q2, we know that the volume of the cone is 2000cm^3, since v=1/3 pi*r^2h, we can make h the subject => h=6000/(r^2*pi). Sub that into q1 and you'll get q2