Question

Find the area of the surface.

The part of the plane \(x + 2y + 3z = 1\) that lies inside the cylinder \({x^2} + {y^2} = 3\) .

Short answer

The required surface area is \(\pi \sqrt {14} \) .

Step-by-step solution

Given

The equation of the plane is\(x + 2y + 3z = 1\)

That is\(z = \frac{1}{3}\left( {1 - x - 2y} \right)\)

The equation of cylinder is\({x^2} + {y^2} = 3\)

The objective is to find area of the surface which is part of the plane that lies inside the cylinder.

FormulasUse the following formulas:

1. The surface area of an equation \(z = g\left( {x,y} \right)\) is ,\(\iint\limits_{D}{\sqrt{{{\left( \frac{\partial z}{\partial x} \right)}^{2}}+{{\left( \frac{\partial z}{\partial y} \right)}^{2}}+1}}dA\)

where the\(D\)is the domain of integration.

1. The fundamental theorem of calculus is\(\int_a^b {f\left( x \right)dx = F\left( b \right) - F\left( a \right)} \),

where \(F\left( x \right)\) is the anti-derivative of \(f\left( x \right)\) .

Finding the domain

First, determine the domain of integration and calculate the necessary partial derivatives.

Since the plane lies inside the cylinder\({x^2} + {y^2} = 3\), the domain of integration in polar coordinates is\(D = \left\{ {\left( {r,\theta } \right)|0 \le r \le \sqrt 3 ,0 \le \theta \le 2\pi } \right\}\)

The partial derivatives of\(z = \frac{1}{3}\left( {1 - x - 2y} \right)\)are,

\(\frac{{\partial z}}{{\partial x}} = - \frac{1}{3},\frac{{\partial z}}{{\partial y}} = - \frac{2}{3}\)

Calculating surface area

The surface area is,

\(\begin{align} & \iint\limits_{D}{\sqrt{{{\left( \frac{\partial z}{\partial x} \right)}^{2}}+{{\left( \frac{\partial z}{\partial y} \right)}^{2}}+1}dA}=\iint\limits_{D}{\sqrt{{{\left( -\frac{1}{3} \right)}^{2}}+{{\left( -\frac{2}{3} \right)}^{2}}+1}dA} \\ & =\iint\limits_{D}{\sqrt{\frac{14}{9}}dA} \\ & =\frac{\sqrt{14}}{3}\int_{0}^{2\pi }{\int_{0}^{\sqrt{3}}{rdrd\theta }} \end{align}\)

Final answer

Simplify further.

\(\begin{align} & \iint\limits_{D}{\sqrt{{{\left( \frac{\partial z}{\partial x} \right)}^{2}}+{{\left( \frac{\partial z}{\partial y} \right)}^{2}}+1}dA}=\frac{\sqrt{14}}{3}\int_{0}^{2\pi }{\int_{0}^{\sqrt{3}}{rdrd\theta }} \\ & =\frac{\sqrt{14}}{3}\left( 2\pi \right)\int_{0}^{\sqrt{3}}{rdr} \\ & =\frac{\sqrt{14}}{3}\int_{0}^{2\pi }{d\theta }\int_{0}^{\sqrt{3}}{rdr} \\ & =\frac{\sqrt{14}}{3}\left( 2\pi \right)\left[ \frac{1}{2}{{r}^{2}} \right]_{r=0}^{r=\sqrt{3}} \end{align}\)

\(\begin{align} & \iint\limits_{D}{\sqrt{{{\left( \frac{\partial z}{\partial x} \right)}^{2}}+{{\left( \frac{\partial z}{\partial y} \right)}^{2}}+1}dA}=\frac{\sqrt{14}}{3}\left( 2\pi \right)\left[ \frac{3}{2} \right] \\ & =\pi \sqrt{14} \end{align}\)

Therefore, the surface area is x\(\pi \sqrt{14}\)