# Volume of a solid

iceman
Hi,
I need help on this problem which is giving me a few headaches...!!!!

here goes..

Show that the volume of the solid bounded by the coordinate planes and the plane tangent to the portion of the surface xyz = k, k>0, in the first octant does not depend on the point of tangency.

Your help will be much appreciated.

OK, i think i have the answer. i make it 9k/2. how much help do you want?

my first hint: the normal to that surface can be found by taking the gradient.

iceman
Hi lethe,

I am totally lost on this question to be honest and I can't seem to work out what to do here to solve it. I would really appreciate it if you could explain step by step what you are doing so I can understand how you came to your conclusion and your answer.

and also your hint: the normal to that surface can be found by taking the gradient.

How would you go about solving this?

Your help will be greatly appreciated.
Thanks.

OK:

step 1: the gradient of xyz - k gives you the normal vector to the surface.

step 2: the equation for a plane with normal vector n is n*(x-x0)=0

so the equation for the tangent plane at x0 is y0z0(x-x0)+x0z0(y-y0)+x0y0(z-z0)=0

or

x/x0 + y/y0 + z/z0 = 3

step 3: find the three coordinate intercepts of this plane by plugging in x=y=0 and get z=30, then x=z=0 and get y=3y0, and x=3x0

step 4: calculate the volume. it is a right pyramid, the base has legs 3x0 and 3y0, so the area of the base is 9x0y0/2. the area for a pyramid is 1/3*Base*height, so this is 9x0y0z0/2, but since x0 is on the surface, x0y0z0 = k, and we get 9k/2 for the volume

iceman
Hey thanks for your help lethe!

Regards
Iceman