   Chapter 5.2, Problem 55E

Chapter
Section
Textbook Problem

# Find the volume of the described solid S.The base of S is an elliptical region with boundary curve 9 x 2 + 4 y 2 = 36. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.

To determine

To find:

The volume of solid S where base of S is an elliptical region with boundary curve 9x2+4y2=36. Also, where cross sections are perpendicular to the  x-axis are isosceles right triangles with hypotenuse in the base.

Explanation

1) Concept:

Definition of volume:

Let S be a solid that lies between x=a and  x=b. If the cross sectional area of S in the plane Px, through x and perpendicular to the x-axis, is  A(x), where A is continuous function, then the volume of S is

V=limni=1nAxi*x=abAxdx

2) Given:

Equation of base curve is 9x2+4y2=36.

3) Calculations:

The given equation of base curve (ellipse) is 9x2+4y2=36.

That is x24+y29=1 Solving for y,

y=±1236-9x2

The top half of the ellipse is with a positive square root

y=1236-9x2

Consider a cross section at x, it is an isosceles triangle. The hypotenuse is a chord joining diametrically opposite point of ellipse. From the diagram, it is clear that there is symmetry between upper and lower part of the x-axis. Thus the hypotenuse at x is given by

hx=2·1236-9x2=36-9x2

Now this is the hypotenuse h of an isosceles right triangle.

By Pythagorean Theorem,

The side length s of the triangle is

s2+s2=h2

That is,s=h2

Then, the area of isosceles right triangle is A=12side2=12s2=h24

The volume of the solid is the sum of areas of each triangle

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