Question: The diagram shows the region enclosed by the parts of the line $y = x/2$ and $x=3$ and the part of the hyperbola $x^2-y^2=12$. The area is rotated about the $y$ - axis to form a solid. Find the volume of the solid.
Please help, I'm kinda stuck and help would be much appreciated. Thank you so much :)
the second diagram is the actual diagram after redrawing it, thanks to the individual that informed me
On
The $x$-intercept of the hyperbola is: $x = 2\sqrt{3}$. We can split it into $2$ solids, and the first one goes from $x = 3 \to x = 2\sqrt{3}$, while the second goes from $x = 2\sqrt{3} \to x=4$. We use Shell method.
$V = \displaystyle \int_{3}^{2\sqrt{3}} 2\pi x\cdot \dfrac{x}{2}dx + \displaystyle \int_{2\sqrt{3}}^{4} 2\pi x\cdot \left(\dfrac{x}{2} - \sqrt{x^2-12}\right)dx$
Inner Radius = $3$ and $2y$
Outer Radius = $\sqrt{(12+y^2)}$
Points of intersection = $(0,\frac{3}{2})$ and $(\frac{3}{2},2)$
Volume Enclosed by the region $=\int_a^{b} \pi\left((Outer radius)^2 - (Inner Radius)^2\right)dy$
$$=\int_0^{\frac{3}{2}} \pi\left(12+y^2 - 3^2\right)dy+\int_{\frac{3}{2}}^{2} \pi\left(12+y^2 - 4y^2\right)dy$$