Laplace Transformation Introduction
Comparison Test

Example 1.

Suppose we would like to find the surface area of z=\frac{2}{3}% \left( x^{3/2}+y^{3/2}\right) that lies above the region bounded by the x
-axis, y-axis and the line 3x+y=3. We know the intersection of the line % 3x+y=3 and the x-axis provides x=1 and when 0\leq x\leq 1, we have % 0\leq y\leq 3-3x. Therefore, we write

    \begin{eqnarray*} A &=&\int_{0}^{1}\int_{0}^{3-3x}\sqrt{\left( \frac{\partial z}{\partial x}% \right) ^{2}+\left( \frac{\partial z}{\partial y}\right) ^{2}+1}dydx \\ &=&\int_{0}^{1}\int_{0}^{3-3x}\sqrt{\left( \sqrt{x}\right) ^{2}+\left( \sqrt{% y}\right) ^{2}+1}dydx \\ &=&\int_{0}^{1}\int_{0}^{3-3x}\sqrt{x+y+1}dydx \end{eqnarray*}

Now evaluate this integral.

Example 2.

Suppose we want to find the surface area of the cone z=\sqrt{% x^{2}+y^{2}} inside the cylinder x^{2}+y^{2}=2x. In this case, the region
that we integrate over is a circle since

    \begin{eqnarray*} x^{2}+y^{2} &=&2x \\ r^{2} &=&2r\cos \theta \\ r &=&2\cos \theta \end{eqnarray*}

Now we write

    \begin{eqnarray*} \frac{\partial z}{\partial x} &=&\frac{x}{\sqrt{x^{2}+y^{2}}} \\ \frac{\partial z}{\partial y} &=&\frac{y}{\sqrt{x^{2}+y^{2}}} \\ \left( \frac{\partial z}{\partial x}\right) ^{2}+\left( \frac{\partial z}{% \partial y}\right) ^{2}+1 &=&1+1 \\ A &=&\int \int_{R}\sqrt{\left( \frac{\partial z}{\partial x}\right) ^{2}+\left( \frac{\partial z}{\partial y}\right) ^{2}+1}dydx \\ &=&\int_{-\pi /2}^{\pi /2}\int_{0}^{2\cos \theta }\sqrt{2}rdrd\theta \end{eqnarray*}

Now evaluate this integral.

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