Question

Find the volume of the solid via cross-sections?

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`

Answer 1

`36 \ \ \ \ \ "units"^3`

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 `x`-value we have:

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

So for that arbitrary `x`-value we have the associated `y`-coordinates `y_1 , y_2` as marked on the image:

`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`