The base of S is the region enclosed by the parabola $y = 9 − 9x^2$ and the X – axis. Cross-sections perpendicular to the X – axis are isosceles triangles with height equal to the base.
[Math] Finding volume of enclosed region
areacalculusconic sectionsvolume
Related Solutions
$$
S : \frac{x^2}{4} + \frac{y^2}{9} = 1
$$
Because the cross sections are isosceles right triangles with hypotenuse in the base, the length of the hypotenuse is $ 6\sqrt{1-\frac{x^2}{4}}$, which means that the area of the cross section is $9(1-\frac{x^2}{4})$
$$ \implies V(x) = \int\limits_{-2}^{2}9(1-\frac{x^2}{4}) \mathrm{d}x $$
That integral should be easy.
Yes, your answer is fine!
This is because the area of an isosceles right triangle with hypotenuse on the base has area equal to $\displaystyle \frac{1}{2}h^2$, and when isolating $y$, you multiply the length of the hypotenuse by $2$ to compensate for the entire ellipse.
Best Answer
You want to take the volume as $$V=\int_a^bA(x)\, dx,$$
where $A(x)$ is a typical area slice. I think from comments you can see that $$A(x)=\frac{1}{2}BH,$$
where $B$ is the base of the triangle and $H$ is the height. Thus $$A(x)=\frac{1}{2}\left(9-9x^2\right)\left(9-9x^2\right).$$
It helps to sketch these things to see what is going on. Learn how to sketch 3 dimensional solids and your life will be much more enjoyable. Next you want the limits of integration. They are the the zeroes of the function $y$, this being where the function meets the $x$ axis, and trivially found. See if $$V=\int_{-1}^{1}\frac{1}{2}\left(9-9x^2\right)^2\, dx$$ meets with your mathematical sensibilities.