# How do you find the volume of the solid with base region bounded by the curve 9x^2+4y^2=36 if cross sections perpendicular to the x-axis are isosceles right triangles with hypotenuse on the base?

Sep 12, 2014

The volume of the solid can be found by
$V = \frac{1}{4} {\int}_{- 2}^{2} \left(36 - 9 {x}^{2}\right) \mathrm{dx} = 24$.

Let us look at some details.
If we rewrite $9 {x}^{2} + 4 {y}^{2} = 36$ as (by dividing by 36)
${x}^{2} / {2}^{2} + {y}^{2} / {3}^{2} = 1$,
we realize that it is en ellipse that spans from $x = - 2$ to $x = 2$.

By solving for $y$, we have
$y = \pm \frac{1}{2} \sqrt{36 - 9 {x}^{2}}$
Since the area $A \left(x\right)$ of the cross-sections can be expressed as

$A \left(x\right) = \frac{1}{2}$(base)(height)
$= \frac{1}{2} \cdot \sqrt{36 - 9 {x}^{2}} \cdot \frac{1}{2} \sqrt{36 - 9 {x}^{2}}$
$= \frac{1}{4} \left(36 - 9 {x}^{2}\right)$

So, the volume of the solid can be found by
$V = \frac{1}{4} {\int}_{- 2}^{2} \left(36 - 9 {x}^{2}\right) \mathrm{dx}$
$= \frac{1}{2} {\int}_{0}^{2} \left(36 - 9 {x}^{2}\right) \mathrm{dx}$
$= \frac{1}{2} {\left[36 x - 3 {x}^{3}\right]}_{0}^{2} = \frac{1}{2} \left(72 - 24\right) = 24$