Question

The base of S is an elliptical region with boundary curve 9x^2+4y^2=36. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base. Find the volume of the described solid S?

Answer 1

Ok.. I'll try to explain it the best way I can, but it's a bit hard since this is 3D stuff, diagrams would be hard.
This is a long one.. sorry for that. But i tried to make it clear. Hope it is.

Well, first, I guess you understand why
\[ 9x^2 + 4y^2 = 36 \]
Comes to represent an ellipse.
However, believe it or not, the real shape of the base doesn't really matter in this case, only the fact that we can get a matching y's for given 'x'.

So what really happens here.
We have a 3D shape with a base with a boundary represented by the formula above.
We are told that "Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base".

In simpler words, this means that if we cut our shape along the X axis to infinitely thin layers, we get points that are all on the same plane.
Those points make some geometrical shape, and if we can calculate all those layers areas we could actually calculate the whole object's volume as it is constructed from all of those.

In our case, those thin layers' shapes are told to be isosceles right triangles, so their hypotenuse is actually apart of the base (!). Which also mean, the right angle is in front of the base.
Since we are slicing upon the X axis, it means the hypotenuse's length is the distance between the two points of the base's boundary. It should be easy enough to see this is the difference between the 2 y's matching for that layer's x.
Let's represent y with x and find the hypotenuse's length:
\[ 9x^2 + 4y^2 = 36 \\
4y^2 = 36 - 9x^2 \; // \sqrt{} \\
2y = \pm \sqrt{36 - 9x^2} = \text{hypotenuse_length}\\
y = \pm \frac{ \sqrt{36 - 9x^2} }{2} \]
Another thing it is important to notice is that if we take any one of those triangles and inscribe them in a circle, since we have inscribed angle of 90, that means the side in front of it (the hypotenuse) is twice the radius of that circle.
But since our triangle is isosceles it also means that the height from the right angle to the hypotenuse also bisectors it.
Since the hypotenuse is the diameter, the middle of it is the center of the circle, and that means that the height is in same length as half of it. basically:
\[ h = \frac{\text{hypotenuse}}{2} = \frac{2y}{2} = y \]

Therefore, the area of that triangle is:
\[ \text{layer_area} = \frac{ h \cdot \text{hypotenuse} }{2} = \frac{y \cdot 2y}{2} = y^2 \\
y = \pm \frac{ \sqrt{36 - 9x^2} }{2} \quad \implies \quad y^2 = \frac{36 - 9x^2}{2} \]
Now, we wanna sum up all those layers for any X in our base.
For that, we need to know at in what range of X values the base is.
Since this is an ellipse those values would be the ones with biggest X possible, and by the base boundary above it means lowest y as possible so:
\[ 9 \cdot x^2 + 4\cdot 0 = 36 \;// \sqrt{}\\
3 \cdot x = \pm 6 \;// \div 3 \\
x = \pm 2 = \text{edge_x_values} \]
Since this is an ellipse we can make it easier and calculate only half of it, and then multiply by 2, as i'm about to do. That means I would use only one of the X edge values, and sum all layers from that to the middle (x = 0)
So now let's integrate this:
\[
\text{volume} = 2 \cdot \int^2_0 y^2 dx = 2 \cdot \int^2_0 \frac{36 - 9x^2}{2} dx = \\
= \huge( \normalsize \frac{36}{2}x -\frac{9}{2} \cdot \frac{x^3}{3} \Huge)|^{\normalsize 2}_{\normalsize 0} \normalsize = \\
= \huge( \normalsize 18x -\frac{3x^3}{2} \Huge)|^{\normalsize 2}_{\normalsize 0} \normalsize = \\
= (18 \cdot 2 - \frac{3 \cdot 2^3}{2}) - (18 \cdot 0 - \frac{3 \cdot 0^3}{2}) = \\
= 36 - 3 \cdot 4 = 36 - 12 = 24
\]
Hope I got it right (and clear..)