Taking the area of isosceles cross sections of an ellipse The base of S is an elliptical region with boundary curve $25x^2 + 4y^2 = 100$. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.
I have gotten the equation of the ellipse in terms of y:
$y=+\frac52\times\sqrt{4-x^2}$ and $y=-\frac52\times\sqrt{4-x^2}$
or, 
$$y=\sqrt{\frac{100-25x^2}{4}}$$ and $$y=-\sqrt{\frac{100-25x^2}{4}}$$
Where do I go from here?
 A: I am assuming that you want to find the volume of the solid created by taking these cross sections. The basic approach to really any of these problems is to find an area formula $A(x)$ or $A(y)$ depending on what you plan on integrating, and the limits of integration. At that point, you will hopefully have a nice integral to work with.
The trick here is that the hypotenuse of these triangles is running from the "top" of the ellipse perpendicular to the $x$ axis, to be "bottom" of the ellipse. The fact that the hypotenuse runs in this direction is important, as it indicates that we want to take cross sections with a thickness $dx$.
To find the area of this solid, we will need the length of the hypotenuse, which will be $2\sqrt{\frac{100-25x^2}{4}}$ (times $2$ because of symmetry), and will yield the length of the hypotenuse at a point. However, what we really need is the length of the other sides.
Note that these are isosceles right triangles which have angles $45^\circ-45^\circ-90^\circ$. The length of these sides have ratios of $1:1:\sqrt2$. The area of these cross sections will be $A=\frac{a^2}{2}$ where $a$ is the length of the sides. Using the pythagorean theorem, we get: $$a^2+a^2=(2\sqrt{\frac{100-25x^2}{4}})^2$$ $$2a^2=(2\sqrt{\frac{100-25x^2}{4}})^2$$ $$\sqrt2a=2\sqrt{\frac{100-25x^2}{4}}$$ $$a=\sqrt{\frac{100-25x^2}{2}}$$
Plugging this into our area equation, we get $$A=\frac{100-25x^2}{4}$$
To find the volume, integrate from $x=-2$ to $x=2$, the leftmost and rightmost points of the base. This will yield the volume of the solid.
$$V=\int_{-2}^{2}{\frac{100-25x^2}{4}dx}=\frac{1}{4}\int_{-2}^{2}{100-25x^2dx}=\left.\frac{1}{4}(100x-\frac{25x^3}{3})\right|_{-2}^{2}=\left.\frac{1}{2}(100x-\frac{25x^3}{3})\right|_{0}^{2}=\frac{200}{3}$$
