some integration question (1 Viewer)

[Deem_Skills]

is this correct so far???

 

[Nedom]

View attachment 36911
is this correct so far???

Do you have the final answer? Not sure if I did it correctly, cause I didn't do your method as I interpreted it differently.

Extra note: Kinda nitpicky, but still some advice is that you should format your w/o so that it is clearer/cleaner, as on your page lots of the stuff is just bunched together, and it's a bit hard to go through. (I can read it, but still can better, I know there are people with really bad handwriting and etc., but better formatting is always easier to look through for a marker)

 

[Deem_Skills]

Do you have the final answer? Not sure if I did it correctly, cause I didn't do your method as I interpreted it differently.

Extra note: Kinda nitpicky, but still some advice is that you should format your w/o so that it is clearer/cleaner, as on your page lots of the stuff is just bunched together, and it's a bit hard to go through. (I can read it, but still can better, I know there are people with really bad handwriting and etc., but better formatting is always easier to look through for a marker)

ohh thx I didn't realise that h was meant to represent the cone height that helps and thanks for the extra note. Answer is πh/3 (R^2 + Rr + r^2)

 

[Nedom]

ohh thx I didn't realise that h was meant to represent the cone height that helps and thanks for the extra note. Answer is πh/3 (R^2 + Rr + r^2)

If you get the correct solution can you post it, cause I don't know where I went wrong.

 

[yanujw]

A more simple method is to consider a linear equation that passes through (0,r) and (h,R). If you rotate it about the x-axis, with boundaries of 0 and h, it will produce the frustum with the dimensions and variables required.

Use the cartesian equation of a line to show that the function would be

Then, the volume is given by

Which you can evaluate to get the result required