The base of S is an elliptical region with boundary curve $25x^2+16y^2=400$. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.
i got 400/3
my work http://puu.sh/6BPTe
is this correct?
[Math] Find the volume of a 3-dimensional body
calculus
Related Solutions
The geometry of this is simple. One parabola faces out from the $y$ axis with vertex at the origin, the other has vertex at $x=3$ and faces opposite. The parabolae intersect at $x=1$, so the cross-sections are bounded by the first parabola for $x \in [0,1]$ and the other for $x \in [1,3]$.
The area of an equilateral triangle of side $s$ is $\sqrt{3} s^2/4$. When bounded within the curves, the side of a triangle at $x$ is $2 y(x)$. The volume element is then $\sqrt{3} ([2 y(x)]^2/4) = \sqrt{3} y^2$. The volume is then
$$\sqrt{3} \int_0^3 dx \: y(x)^2 = \sqrt{3} \int_0^1 dx \: x + \sqrt{3} \int_1^3 dx \: \frac{1}{2} (3-x)$$
I trust you can evaluate these.
$$
S : \frac{x^2}{4} + \frac{y^2}{9} = 1
$$
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.
Best Answer
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.