The graph above shows the base of an object. The shape is outlined in blue on a coordinate grid, spanning from x = 1 to x = 9 and y = -5 to y = 4.To describe the shape:- It starts at point (1, -1), goes up diagonally to (4, 4).- From (4, 4), it goes horizontally to (8, 4).- Then, it moves diagonally down to (9, -5).- Finally, it connects back horizontally to (1, -5) and returns to the starting point (1, -1).Compute the exact value of the volume of the object, given that cross sections (perpendicular to the base) are isosceles right triangles with their hypotenuse in the base.$ V = $\[ 48 \](Note: The answer 48 is marked as incorrect.)

Answered: The graph above shows the base of an…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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The graph above shows the base of an object. The shape is outlined in blue on a coordinate grid, spanning from x = 1 to x = 9 and y = -5 to y = 4.

To describe the shape:

- It starts at point (1, -1), goes up diagonally to (4, 4).
- From (4, 4), it goes horizontally to (8, 4).
- Then, it moves diagonally down to (9, -5).
- Finally, it connects back horizontally to (1, -5) and returns to the starting point (1, -1).

Compute the exact value of the volume of the object, given that cross sections (perpendicular to the base) are isosceles right triangles with their hypotenuse in the base.

$ V = $
\[ 48 \]
(Note: The answer 48 is marked as incorrect.)

Expert Solution

Step 1: Analysis and Introduction

Given Information:

The graph shows the base of the object.

It is as isosceles right triangle from $x equals 0 space to space x equals 6$ and the rectangle formed as the hypotenuse of the isosceles triangle as the side of it from $x equals 6 space to space x equals 8$ .

To find:

The volume of the solid when cross-section perpendicular to its base.

Concept used:

Volume of the object is determined as $integral subscript a superscript b A open parentheses x close parentheses d x$ cubic units, where $A open parentheses x close parentheses$ is the area of the cross-section.

Area of the rectangle is the product of the length and the width.

Area of the triangle is the half of the product of the height and its base.

The hypotenuse of the isosceles right triangle is $square root of 2 a$ , where $a$ is the base (height) of the triangle.

Integration formula:

$integral x to the power of n d x equals fraction numerator x to the power of n plus 1 end exponent over denominator n plus 1 end fraction plus c$ , for all $n not equal to negative 1$ . Here, $c$ is an integrating formula.

$integral subscript a superscript b f open parentheses x close parentheses d x equals F open parentheses b close parentheses minus F open parentheses a close parentheses$ , where $F apostrophe open parentheses x close parentheses equals f open parentheses x close parentheses$ .