# Thread: surface area of portion of plane

1. ## surface area of portion of plane

find the surface area of that portion of the plane z=1 + 2x - 3y that lies above the xy-plane

i don't understand how to set up the integral for this and i think it is easier to do it one way either dxdy or dydx but i am so lost...can any one help me????

2. Unless you are talking about the portion that lies above the 2nd quadrant in the xy plane the surface area is infinite.

I think that this is what you are looking for.

If so this will work

$z=1+2x-3y$

using the formula

$\iint_S dS=\iint_D\sqrt{1+\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}}dA=\iint_D\sqrt{1+(2)^2+(-3)^2}dA=\sqrt{14}\iint_D dA$

From here the calculation of the integral can be done by letting y go from $y=0$ to $y=\frac{2}{3}x+\frac{1}{3}$ and x from $x=-\frac{1}{2}$ to $x=0$

Or you can notice the region in the 2nd quadrant is a triangle with a base of $\frac{1}{2}$ and a height of $\frac{1}{3}$

$A=\frac{1}{2}bh=\left( \frac{1}{2} \right)\left( \frac{1}{2} \right)\left( \frac{1}{3} \right)=\left( \frac{1}{12} \right)$

So finally we get

$\sqrt{14}\iint_D dA=\sqrt{14}\left( \frac{1}{12}\right)=\frac{\sqrt{15}}{12}$

3. Empty done a fine job, but if I may show you one other little thing about these.

We can set it up using dydx, dzdy, or dzdx and get the same thing.

For dzdy

$\int_{0}^{\frac{1}{3}}\int_{0}^{1-3y}\frac{\sqrt{14}}{2}dzdy=\frac{\sqrt{14}}{12}$

What we do is solve the original equation for x, thus getting

$x=\frac{z+3y-1}{2}$. Now, if we set x and z = 0 and solve for z we get the limit for z. Which is 0 to $z=1-3y$

Of course, to get the y limit, set x and z=0 and solve for y and we see y=1/3.

To do dzdx, it works the same way and we get:

$\int_{-\frac{1}{2}}^{0}\int_{0}^{2x+1}\frac{\sqrt{14}}{3} dzdx=\frac{\sqrt{14}}{12}$

See what I mean?.