# Find the volume of the solid via cross-sections?

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*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.*

**Answer key:** #V=36#

*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.*

**Answer key:**

##### 1 Answer

#### Explanation:

Consider a vertical view of the base of the object.

The grey shaded area represents a top view of the right angled triangle cross section. In order to find the volume of the solid we seek the volume of a generic cross sectional triangular "slice" and integrate over the entire base (the circle)

The equation of the circle is:

# x^2 + y^2 = 3^2 #

So for some arbitrary

# y^2=9-x^2 #

# :. y = +-sqrt(9-x^2) #

So for that arbitrary

# y_1 = +sqrt(9-x^2) #

# y_2 = -sqrt(9-x^2) #

Thus, the length of the base of an arbitrary cross sectional triangular slice is:

# l = y_1 - y_2 #

# \ \ = sqrt(9-x^2) - (-sqrt(9-x^2) ) #

# \ \ = 2sqrt(9-x^2) #

The following depicts a side view of the triangular slice.

![Steve M]

Thus the Area of an arbitrary cross sectional triangular slice is:

# A_("slice") = 1/2 xx "base" xx "height" #

# " " = 1/2( 2sqrt(9-x^2) ) ( sqrt(9-x^2) ) #

# " " = 9-x^2 #

Finally, the volume of the entire solid is the sum of those arbitrary cross sectional slices over the circular base:

# V = sum_("circle") lim_(delta x rarr 0) A_("slice") delta x #

# \ \ \ = int_(-3)^(3) 9-x^2 dx #

# \ \ \ = [9x-x^3/3]_(-3)^(3) #

# \ \ \ = (27-9) - (-27+9) #

# \ \ \ = 27-9+27-9#

# \ \ \ = 36 #