Volume of solids

Volume of solids

Postby ganeshm » Sat Oct 03, 2020 11:12 pm

Consider the 3d rectangular coordinate system.

I have a semicircle on the x axis with radius=1, drawn between x=-1 and x=+1. I know that the area under this curve is [tex]\pi[/tex]/2

I draw another semicircle drawn on the y axis with radius=1, between y=-1 and y=+1. The area under this curve too is [tex]\pi[/tex]/2

By one of the rules of multivariable calculus, multiplying these two areas should give me the volume under a hemisphere, but I know it does not.

Where am I going wrong?
ganeshm
 
Posts: 1
Joined: Sat Oct 03, 2020 11:02 pm
Reputation: 0

Re: Volume of solids

Postby Guest » Sat Oct 10, 2020 2:09 pm

ganeshm wrote:Consider the 3d rectangular coordinate system.

I have a semicircle on the x axis with radius=1, drawn between x=-1 and x=+1. I know that the area under this curve is [tex]\pi[/tex]/2

I draw another semicircle drawn on the y axis with radius=1, between y=-1 and y=+1. The area under this curve too is [tex]\pi[/tex]/2

By one of the rules of multivariable calculus, multiplying these two areas should give me the volume under a hemisphere, but I know it does not.

Which "rule of multivariable calculus is that? I've never heard of such a rule!

It should be clear that if you multiply two "areas", that have units of "distance squared" ("square meters", "square feet") the result would have units of "distance to the fourth power" while volume has units of "distance cubed" ("cubic meters", "cubic feet").
Where am I going wrong?

I suspect that you are misremembering the "rules of multivariable calculus". Can you tell where you got this "rule"
Guest
 


Return to Calculus - integrals, lim, functions



Who is online

Users browsing this forum: No registered users and 5 guests