The area bounded above by the line $y = 3$, below by the line $y = 0$, on the left by the y-axis and on the right by an arc of the hyperbola $9x^2 – 16y^2 = 144$ is rotated around the x-axis. Find the volume of the solid generated.
I first found the intersection point between the hyperbola and the line $y = 3$ to be at $x = 4\sqrt{2}$. From there, I first found the area from $x = 0$ to $x = 4$ and then from $x = 4$ to $x = 4\sqrt{2}$. The following is my work:
From $x = 0$ to $x = 4$:
$$\int_0^4 \pi(3)^2 dx = 36\pi$$
From $x = 4$ to $x = 4\sqrt{2}$:
$$\int_4^{4\sqrt{2}} \biggl(\frac{9x^2}{16} – 9\biggl)\pi dx = 24\pi\biggl(1 – \frac{\sqrt{2}}{2}\biggl)$$
I then added both of these areas. However, in the book where I was reading this problem, the answer is supposed to be $24\pi\biggl(2\sqrt{2} – 1\biggl)$. I'm new to solids of revolution and so far I've only learned how to find the volume using the cylindrical method.
If anyone can help me understand what I did wrong or give a hint on how to solve the problem I would appreciate it.
Best Answer
Since $y=3$ lies above the branch of the hyperbola, your second integral should be $$\int_4^{4\sqrt{2}} \left(9-\frac{9x^2-144}{16}\right)\ dx = 48 \sqrt{2} \pi -60 \pi.$$