Recent content by sunglasboy

hmm its still not very clear to me. I am having lots of trouble thinking in 3d..

thanks for the suggestion. So basically my single integral would be- pie[(h-z)r/h]^2 dz evaluated from 0 to h ? Is this correct? Still unsure how to proceed with the double and triple integrals. Thanks

I am currently having trouble solving this problem "Find the volume of a right, circular cone of radius r and height h" a) as a single integral. b) as a double integral. c) as a triple integral. My difficulty lies in setting up the integrals. I usually have trouble with problems that do not...