Surface Area

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AP Calculus BC › Surface Area

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1

Find the surface area of the part of the plane in the first octant.

$\frac{27}{8}$

$\frac{17}{8}$

$\frac{13}{8}$

Explanation

Lets recall the equation of surface area.

$S=\int \int_D \sqrt{[f_x]^2+[f_y]^2+1}\ dA$

Now we need to find all the neccessary equations to be able to evaluate the integral.

$\frac{\partial z}{\partial x}=-2$

$\frac{\partial z}{\partial y}=-2$

We will plug in , into the plane equation in order to get a line that intersects with the z axis.

$4x+4y+2(0)=6\rightarrow y=\frac{3}{2}-x$

Now we are going to set , in the previous equation and solve for .

$0=\frac{3}{2}-x\rightarrow x=\frac{3}{2}$

We now have all the bounds for our double integral

$0\leq x \leq \frac{3}{2}$

$0\leq y \leq \frac{3}{2}-x$

$S=\int_{0}^{\frac{3}{2}} \int_{0}^{\frac{3}{2}-x} \sqrt{[-2]^2+[-2]^2+1} \ dy \ dx$

$S=\int_{0}^{\frac{3}{2}} \int_{0}^{\frac{3}{2}-x} \sqrt{9} \ dy \ dx$

$S=\int_{0}^{\frac{3}{2}} \int_{0}^{\frac{3}{2}-x} 3 \ dy \ dx$

$S=\int_{0}^{\frac{3}{2}} 3y \Big|_{0}^{\frac{3}{2}-x} \ dx$

$S=\int_{0}^{\frac{3}{2}} 3(\frac{3}{2}-x)-0 \ dx$

$S=\int_{0}^{\frac{3}{2}} \frac{9}{2}-3x \ dx$

$S= \frac{9}{2}x-\frac{3}{2}x^2 \Big|_{0}^{\frac{3}{2}}$

$S= \frac{9}{2}(\frac{3}{2})-\frac{3}{2}(\frac{3}{2})^2-0=\frac{27}{8}$

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