Question

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit representation of the surface.

f(x,y,z) = ez; S is the plane z = 8 - x - 2y in the first octant.

Answer 1

Parameterize
`S` by the vector function

`\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k`

so that the normal vector to
`S` is given by

`(\partial\vec r)/(\partial x)*(\partial\vec r)/(\partial y)=\left(\vec\imath+(\partial f)/(\partial x)\,\vec k\right)*\left(\vec\jmath+(\partial f)/(\partial y)\,\vec k\right)=-(\partial f)/(\partial x)\vec\imath-(\partial f)/(\partial y)\vec\jmath+\vec k`

with magnitude

`\left\|(\partial\vec r)/(\partial x)*(\partial\vec r)/(\partial y)\right\|=\sqrt{\left((\partial f)/(\partial x)\right)^2+\left((\partial f)/(\partial y)\right)^2+1}`

In this case, the normal vector is

`(\partial\vec r)/(\partial x)*(\partial\vec r)/(\partial y)=-(\partial(8-x-2y))/(\partial x)\,\vec\imath-(\partial(8-x-2y))/(\partial y)\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k`

with magnitude
`√(1^2+2^2+1^2)=\sqrt6`. The integral of
`f(x,y,z)=e^z` over
`S` is then

`\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^(8-x-2y)\,\mathrm dy\,\mathrm dx`

where
`T` is the region in the
`x,y` plane over which
`S` is defined. In this case, it's the triangle in the plane
`z=0` which we can capture with
`0\le x\le8` and
`0\le y\le\frac{8-x}2`, so that we have

`\displaystyle\sqrt6\iint_Te^(8-x-2y)\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^((8-x)/2)e^(8-x-2y)\,\mathrm dy\,\mathrm dx=\boxed{√(\frac32)(e^8-9)}`