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Q. 22

Expert-verified

Calculus

Found in: Page 1119

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Find the area of S is the portion of the plane with equation x = y + z that lies above the region in the xy-plane that is bounded by y = x, y = 5, y = 10, and the y-axis.

$Hence the area of triangle is = \frac{75 \sqrt{3}}{2}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given

Equation of plane x=y+ z

Step 2. Formula for finding the area of surface

$If a surface S is given by z = f (x, y) for f (x, y) \in D \subset R^{2} Then the surface area of smooth surface is \begin{aligned} \int_{S} d S = \int_{S} \sqrt{{(\frac{\partial z}{\partial x})}^{2} + {(\frac{\partial z}{\partial y})}^{2} + 1 d A} \\ = \iint_{D} \sqrt{{(\frac{\partial z}{\partial x})}^{2} + {(\frac{\partial z}{\partial y})}^{2} + 1 d A \dots \dots (1} \end{aligned}$

Step 3. Finding the partial derivative

$x = y + z t h e n, z = x - y n o w, f i r s t f i n d \begin{aligned} \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (x - y) \\ = 1 \end{aligned} a n d \begin{aligned} \frac{\partial z}{\partial y} = \frac{\partial}{\partial y} (x - y) \\ = - 1 \end{aligned}$

Step 4. Graph

viewed it as a x-simple, then the region of integration will be

$D = \{(x, y) | 0 \leq x \leq y, 5 \leq y \leq 10\}$

Step 5. Finding the area

$T h e a r e a o f s u r f a c e i s \begin{aligned} \int_{S} d S = \iint_{D} \sqrt{{(\frac{\partial z}{\partial x})}^{2} + {(\frac{\partial z}{\partial y})}^{2} + 1} d A \\ = \int_{5}^{10} \int_{0}^{y} \sqrt{1^{2} + - 1^{2} + 1} d x d y \\ = \int_{5}^{10} \int_{0}^{y} \sqrt{3} d x d y \\ = \sqrt{3} \int_{5}^{10} \int_{0}^{y} d x d y \\ = \sqrt{3} \int_{5}^{10} \int_{0}^{y} d x d y \\ = \sqrt{3} \int_{5}^{10} [x]_{0}^{y} d y \\ = \sqrt{3} \int_{5}^{10} [y - 0] d y \\ = \sqrt{3} \int_{5}^{10} y d y \\ = \sqrt{3} \int_{5}^{10} y d y \\ = \sqrt{2} [\frac{y^{2}}{2}]_{5}^{10} \\ = \sqrt{2} [\frac{10^{2}}{2} - \frac{5^{2}}{2}] \\ = \frac{75 \sqrt{3}}{2} \\ Hence the area of triangle is = \frac{75 \sqrt{3}}{2} \end{aligned}$

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