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Find the area of the following surfaces using a parametric description of the surface. a. The...

Question:

Find the area of the following surfaces using a parametric description of the surface.

a. The plane z = 3 - x - 3y in the first octant (where x, y, and z are positive)

b. The hemisphere {eq}x^2 + y^2 + z^2 = 100 {/eq} for {eq}z \geq 0. {/eq}

Integration

A mathematical quantity that represents the combination of the derivative of the function or an equation is known as integration. It helps in the physics discipline to find the volume of the curve.

Answer and Explanation:

(a)

Given Data:

  • The plane is: {eq}z = 3 - x - 3y \cdots\cdots\rm{(I)} {/eq}


Differentiate the expression (I) with respect to x

{eq}\begin{align*} \dfrac{{dz}}{{dx}} &= \dfrac{{d\left( {3 - x - 3y} \right)}}{{dx}}\\ &= - 1 \end{align*} {/eq}


Differentiate the expression (I) with respect to y

{eq}\begin{align*} \dfrac{{dz}}{{dy}} &= \dfrac{{d\left( {3 - x - 3y} \right)}}{{dx}}\\ &= - 3 \end{align*} {/eq}


For the {eq}z = 0 {/eq} expression (I) is

{eq}\begin{align*} 0 &= 3 - x - 3y\\ x + 3y &= 3 \end{align*} {/eq}


The value of y from above expression

{eq}\begin{align*} x + 3y &= 3\\ y &= 1 - \dfrac{x}{3} \end{align*} {/eq}


The domain in x-y plane is

{eq}\begin{align*} 0 &\le x \le 3\\ 0 &\le y \le 1 - \dfrac{x}{3} \end{align*} {/eq}


The expression for area of surface by integration is

{eq}S = \int_0^3 {\int_0^{1 - \dfrac{x}{3}} {\sqrt {1 + {{\left( {\dfrac{{dz}}{{dx}}} \right)}^2} + {{\left( {\dfrac{{dz}}{{dy}}} \right)}^2}} } } dydx {/eq}


Substitute and solve the above expression

{eq}\begin{align*} S &= \int_0^3 {\int_0^{1 - \dfrac{x}{3}} {\sqrt {1 + {{\left( { - 1} \right)}^2} + {{\left( { - 3} \right)}^2}} } } dydx\\ &= \int_0^3 {\int_0^{1 - \dfrac{x}{3}} {\sqrt {11} } } dydx\\ &= \sqrt {11} \int_0^3 {\left[ y \right]} _0^{1 - \dfrac{x}{3}}dx\\ &= \sqrt {11} \int_0^3 {\left[ {\left( {1 - \dfrac{x}{3}} \right) - 0} \right]} dx\\ &= \sqrt {11} \int_0^3 {\left[ {1 - \dfrac{x}{3}} \right]} dx \end{align*} {/eq}


Integrate the above expression with respect to x

{eq}\begin{align*} S &= \sqrt {11} \int_0^3 {\left[ {1 - \dfrac{x}{3}} \right]} dx\\ &= \sqrt {11} \int_0^3 {\left[ {x - \dfrac{{{x^2}}}{6}} \right]} _0^3\\ &= \sqrt {11} \left[ {\left( {3 - \dfrac{{{{\left( 3 \right)}^2}}}{6}} \right) - 0} \right]\\ &= \dfrac{{3\sqrt {11} }}{2} \end{align*} {/eq}


Thus the surface area is {eq}\dfrac{{3\sqrt {11} }}{2} {/eq}


(b)

  • The hemisphere curve is: {eq}{x^2} + {y^2} + {z^2} = 100 {/eq}


The rewrite the hemisphere curve is

{eq}{x^2} + {y^2} + {z^2} = {\left( {10} \right)^2} {/eq}


The radius of curvature in polar coordinate is

{eq}{R^2} = {x^2} + {y^2} {/eq}


Substitute and solve the hemisphere curve expression

{eq}\begin{align*} {R^2} + {z^2} &= 100\\ z &= \sqrt {100 - {R^2}} \end{align*} {/eq}


The domain of the curve is

{eq}\begin{align*} 0 &\le R \le 10\\ 0 &\le z \le \sqrt {100 - {R^2}} \\ 0 &\le \theta \le 2\pi \end{align*} {/eq}


The expression for surface area of the curve by integration is

{eq}S = \int_0^{2\pi } {\int_0^{10} {\int_0^{\sqrt {100 - {R^2}} } {RdzdRd\theta } } } {/eq}


Integrate the above expression with respect to z

{eq}\begin{align*} S &= \int_0^{2\pi } {\int_0^{10} {\left[ z \right]_0^{\sqrt {100 - {R^2}} }RdRd\theta } } \\ &= \int_0^{2\pi } {\int_0^{10} {\left[ {\sqrt {100 - {R^2}} - 0} \right]RdRd\theta } } \\ &= \int_0^{2\pi } {\int_0^{10} {\left[ {\sqrt {100 - {R^2}} } \right]RdRd\theta } } \cdots\cdots\rm{(II)} \end{align*} {/eq}


Let {eq}V = 100 - {R^2} \cdots\cdots\rm{(III)} {/eq}


Differentiate the above expression with respect to {eq}R {/eq}

{eq}\begin{align*} \dfrac{{dV}}{{dR}} &= \dfrac{{d\left( {100 - {R^2}} \right)}}{{dR}}\\ \dfrac{{dV}}{{dR}} &= - 2R\\ dR &= - \dfrac{{dV}}{{2R}} \end{align*} {/eq}


Substitute and solve the expression (III) at {eq}R = 0 {/eq}

{eq}\begin{align*} V &= 100 - {\left( 0 \right)^2}\\ &= 100 \end{align*} {/eq}


Substitute and solve the expression (III) at {eq}R = 10 {/eq}

{eq}\begin{align*} V &= 100 - {\left( {10} \right)^2}\\ &= 0 \end{align*} {/eq}


The domain of the variable {eq}V {/eq} is

{eq}100 \le V \le 0 {/eq}


Substitute and solve the expression (II)

{eq}\begin{align*} S &= \int_0^{2\pi } {\int_{100}^0 {\left[ {\sqrt V } \right]R\left( { - \dfrac{{dV}}{{2R}}} \right)d\theta } } \\ &= - \dfrac{1}{2}\int_0^{2\pi } {\int_{100}^0 {\sqrt V } } dVd\theta \\ &= - \dfrac{1}{2}\int_0^{2\pi } {\dfrac{2}{3}\left[ {{{\left( V \right)}^{\dfrac{3}{2}}}} \right]_{100}^0d\theta } \\ &= - \dfrac{1}{2}\int_0^{2\pi } {\dfrac{2}{3}\left[ {{{\left( 0 \right)}^{\dfrac{3}{2}}} - {{\left( {100} \right)}^{\dfrac{3}{2}}}} \right]_{100}^0d\theta } \\ &= \dfrac{1}{2}\int_0^{2\pi } {\dfrac{2}{3}\left[ {1000} \right]d\theta } \\ &= \dfrac{{1000}}{3}\int_0^{2\pi } {d\theta } \end{align*} {/eq}


Integrate the above expression with respect to {eq}\theta {/eq}

{eq}\begin{align*} S &= \dfrac{{1000}}{3}\left[ \theta \right]_0^{2\pi }\\ &= \dfrac{{1000}}{3}\left[ {2\pi - 0} \right]\\ &= \dfrac{{2000\pi }}{3} \end{align*} {/eq}


Thus the surface area of curve is {eq}\dfrac{{2000\pi }}{3} {/eq}


Learn more about this topic:

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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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