# How to find the volume of a hemisphere on top of a cone

Volume of hemi-sphere = ∫ ∫ ∫ r2 sinθ dr dθ dφ

i thing (r < r < (r + R)cosθ ) ( 0 < θ < 60 = π/6) and ( 0 < φ < 2π)

integral = 2π ∫ ⅓r3 sin θ dθ

= 2π ∫ ⅓ [((r+R)cosθ)3 - r3] sin θ dθ

i don't know how to find volume of hemi-spere

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fresh_42
Mentor
Please do not open more than one thread with the same topic, especially if the two are both ambiguous: with or without cone, what is ##r## needed for and what is ##a## in your other thread. Furthermore, do not delete the homework template, use it! It makes reading a lot easier and if you delete it, it can be viewed as disrespectful to those who are willing to answer.

I closed the other one.

Please do not open more than one thread with the same topic, especially if the two are both ambiguous: with or without cone, what is ##r## needed for and what is ##a## in your other thread. Furthermore, do not delete the homework template, use it! It makes reading a lot easier and if you delete it, it can be viewed as disrespectful to those who are willing to answer.

I closed the other one.
I'm sorry

i have problem about find volume of hemisphere on cone using triple integral. (spherical coordinates)
I do not know the true extent of r (From 0 to ??????)

fresh_42
Mentor
Beside what I've written in the other thread, with the mistakes mentioned and referring to the hemisphere without the cone involved, the final radius is ##R##. You had it almost all, beside that ##\cos \frac{\pi}{2}=0## and ##\cos 0 = 1## you only had to solve ##\int_0^R r^2dr## plus eventually the volume of the cone. I assume that it is a full hemisphere above the cone and the angle of ##30°## refers to the cone alone.

LCKurtz