Question

Find the area of the surface. the part of the plane 3x 5y z = 15 that lies in the first octant

Answer 1

The area is 97.211 sq units.

This part of the plane is a triangle. We can Call it δ. We can find the intercepts by setting two variables to 0 simultaneously; we'd find, for instance, that y=z=0 means 3x=15 i.e., x=3 , so that (3, 0, 0) is one vertex of the triangle. Similarly, we'd find that (0, 5, 0) and (0, 0, 15) are the other two vertices.

Next, we can parameterize the surface by

s(u,v)=$\int\limits^a_b {3(1-u)(1-v),5u(1-v),15v} \, dx$

so surface element is

dS= IIsu*svII =15$\sqrt{42}$(1-v)dudv

Then the area of  is given by the surface integral

$\int\limits^a_b {} \, \int\limits^a_b {dS} \, dx = 15\sqrt[n]{42} \int\limits^a_b {x} \, dx \int\limits^a_b {1-v} \, dudv$=97.211

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