   Chapter 5.R, Problem 23E

Chapter
Section
Textbook Problem

# The base of a solid is a circular disk with radius 3. Find the volume of the solid if parallel cross-sections perpendicular to the base are isosceles right triangles with hypotenuse lying along the base.

To determine

To find:

The volume of the solid S where parallel cross sections perpendicular to the base are isosceles triangles with hypotenuse lying along the base.

Explanation

1) Concept:

Use the definition of volume.

2) 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

3) Calculations:

Fig (a) represents the graph of a circle that is the base of solid S and has radius 3.

The red line y is the hypotenuse of an isosceles right triangle represented in fig (b).

From fig (a), the hypotenuse of the triangle can be found by finding the distance between the top half of the circle and the x-axis and multiplying that by 2.

The equation of circle with radius 3 is

x2+y2=9

Solving for y,

y=9-x2

From fig (a),

Hypotenuse =2y=29-x2.

Now, find the length of the other two sides of an isosceles triangle by Pythagoras theorem.

a2+b2=hypotenuse2

Since it is an isosceles triangle, length of the other two sides is equal.

Therefore, substitute a=b in the above equation

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