Question

Find the volume V of the described solid S. The base of S is an elliptical region with boundary curve 16x2 + 9y2 = 144. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.

Answer 1

In the $x$-$y$ plane, the base has equation(s)

$16x^2+9y^2=144\implies y=\pm\dfrac43\sqrt{9-x^2}$

which is to say, the distance (parallel to the $y$-axis) between the top and the bottom of the ellipse is

$\dfrac43\sqrt{9-x^2}-\left(-\dfrac43\sqrt{9-x^2}\right)=\dfrac83\sqrt{9-x^2}$

so that at any given $x$, the cross-section has a hypotenuse whose length is $\dfrac83\sqrt{9-x^2}$.

The cross-section is an isosceles right triangle, which means the legs occur with the hypotenuse in a ratio of 1 to $\sqrt2$, so that the legs have length $\dfrac8{3\sqrt2}\sqrt{9-x^2}$. Then the area of each cross-section is

$\dfrac12\left(\dfrac8{3\sqrt2}\sqrt{9-x^2}\right)\left(\dfrac8{3\sqrt2}\sqrt{9-x^2}\right)=\dfrac{16}9(9-x^2)$

Then the volume of this solid is

$\displaystyle\frac{16}9\int_{-3}^39-x^2\,\mathrm dx=\boxed{64}$