# 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: $V = 36$

Sep 20, 2017

$36 \setminus \setminus \setminus \setminus \setminus {\text{units}}^{3}$

#### 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:

${y}^{2} = 9 - {x}^{2}$
$\therefore y = \pm \sqrt{9 - {x}^{2}}$

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:

$l = {y}_{1} - {y}_{2}$
$\setminus \setminus = \sqrt{9 - {x}^{2}} - \left(- \sqrt{9 - {x}^{2}}\right)$
$\setminus \setminus = 2 \sqrt{9 - {x}^{2}}$

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"
$\text{ } = \frac{1}{2} \left(2 \sqrt{9 - {x}^{2}}\right) \left(\sqrt{9 - {x}^{2}}\right)$
$\text{ } = 9 - {x}^{2}$

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

$V = {\sum}_{\text{circle") lim_(delta x rarr 0) A_("slice}} \delta x$
$\setminus \setminus \setminus = {\int}_{- 3}^{3} 9 - {x}^{2} \mathrm{dx}$

$\setminus \setminus \setminus = {\left[9 x - {x}^{3} / 3\right]}_{- 3}^{3}$

$\setminus \setminus \setminus = \left(27 - 9\right) - \left(- 27 + 9\right)$

$\setminus \setminus \setminus = 27 - 9 + 27 - 9$

$\setminus \setminus \setminus = 36$