Finding the volume of a solid with limitations

Question

I have a question that is getting me on my nerves: Find the volume of a solid limited by $64x^2 - 4y^2 + 16z^2 = 0$ and $y = 1$. Can anyone help me? Thank you!

Answer 1

So your solid is limited by $y=1$ and $y = 16x^2+4z^2$. you can set this up as a triple integral (in cylindrical coordinates would be ideally easy), bounding $16x^2+4z^2 \le y \le 1$ and integrate over the ellipse in the $xz$-plane...

Answer 2

Okey. I suppose that it's also limited by $y=0$. We have elliptical cone, which volume is $V = \frac{1}{3} \pi a b h$. Where the $h = 1$, because volume is bounded between $y=0$ and $y=1$. To find $a$ and $b$ we need to put $y=1$ to equation:

$$64x^2 + 16z^2 = 4$$

$$16x^2 + 4z^2 = 1$$

Hence we get $a = 1/4$ and $b = 1/2$. Therefore, $V = \frac{\pi}{24}$

Answer 3

Fact: For ellipse $a$ as the length of semi-major axis and $b$ be the length of semi-minor axis, the area of the ellipse is $\pi a b$

Here, given equation $$ y^2 = 16 x^2 + 4 z^2 $$ Let $a,b$ be the semi-minor and semi-major axis of any ellipse lying in level plane, and $a_0, b_0$ be the values of semi-minor and semi-major axis when $y = 1$ Whose level surfaces are ellipse, when $y = 1$ ,$ a_0 = 1/4,b_0 = 1/2$ as ratio of $y$ to $b$ (or $a$) remains constant, at any value of $y$, say $y = k$, $k = 4a$, the region of graph for $y \in [0,1]$ can be subdivided into elliptical disc of thickness $dk$ ,and area $ \pi a b$ , $$ \Rightarrow dV = \pi \cdot a \cdot b \cdot dk =2 \pi a^2 dk = 8 \pi a^2 da$$

because for the ellipse here $\cfrac{b}{a} = 2$. As $k = 4a$ as $k \in [0,1] \Rightarrow a \in [0,1/4]$ $$ \Rightarrow V = \int_0^{1/4} 8 \pi a^2 da = \cfrac{\pi}{24} $$