Does the Volume of a Solid Depend on Point of Tangency?

[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.

 

[lethe]

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.

eg. how you came to your answer of 9k/2.

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.

 

[lethe]

OK:

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

the gradient is (yz,xz,xy)

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