Find the volume of the solid bounded by surfaces.

Question

Find the volume of a body formed by rotation around an axis $ V_{x}$

Figure bounded by the graphs of functions:

$$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$$

$$x^{2}-\frac{y^{2}}{15}=1$$

$$x \geq 1$$ This is my picture.

Answer 1

The volume were talking about has the following plot

We want the volume when that is rotated, so if we rotate the top part around the x-axis with the following formula, it should work. $$ V=\pi\int_a^bf(x)^2\,\mathrm dx $$ Now, to find the formula to put in $f(x)$, that function has to be equal to the smallest of our two functions. In the plot we can see that the purple function is smallest at first, but then it becomes the green one. Therefore we must know their intersection, so we know when to switch between the two functions.

To find their intersection, solve for $y^2/15$ in the first formula, and substitute into the second, then solve for $x$ in the second and obtain that their intersection lies at $$ x=\pm\frac54 $$ We are interested in the positive solution. Now use our volume formula from $1$ to $\frac54$ on the purple formula, and then from $\frac54$ to $5$ which is where th green shape stops. The sum of these two things will then be the desired volume. Finding $f(x)$ for each of these is equivalent to solving for $y$, but since we're looking for $f(x)^2$ we can actually solve for $y^2$ and simply use that: $$ \begin{align} \frac{x^2}{25}+\frac{y^2}9=1&\implies y^2=9-\frac{9 x^2}{25}\\ x^2-\frac{y^2}{15}=1&\implies y^2=15x^2-15 \end{align} $$ Putting that into the formula as previously descriped gives: $$ \begin{align} V&=\pi\int_1^{5/4}15x^2-15\,\mathrm dx+\pi\int_{5/4}^59-\frac{9 x^2}{25}\,\mathrm dx\\ &=20\pi \end{align} $$