Volume of a cylinder in paraboloid.

Question

Find the volume lying inside the cylinder $x^2 + y^2 – 2x = 0$ and outside the paraboloid $x^2 + y^2 = 2z$, while bounded by $xy$-plane.

Answer 1

The equation of the paraboloid is $x^2+y^2 = 2z.$ So it is same as taking a parabola $x^2 = 2z$ where $y = 0$ or taking $y^2 = 2z$ where $x = 0$ and rotating it around $z$-axis. Here, $z \ge 0.$

The cylinder $x^2 + y^2 = 2x$ can be described as a cylinder with radius $1$ and touching $yz$ plane with $ 0 \le x \le 2$ and $ -1 \le y \le 1$ and $z \ge 0$ as it is bound by $xy$ plane on the same side as paraboloid. For z = 0, it is a circle of radius 1 with center at $(1,0)$. You can also write it as $(x-1)^2+y^2 = 1.$ I hope you can now visualize it.

In polar coordinates, you can write it as

$$[{ (r, θ) | − π/2 ≤ θ ≤ π/2, 0 ≤ r ≤ 2\cos θ }]$$

And the height is bound by $z$ value of the paraboloid which is $\frac{1}{2}(x^2+y^2).$

You can integrate the area of the circle (A) between these bounds to find the volume.

$$\iint \frac{1}{2}(x^2+y^2)dA$$

$$\frac{1}{2}\int^{\pi/2}_{-\pi/2}\int^{2\cos\theta}_{0}r^2\cdot r\cdot dr\cdot d\theta$$

$$2\int^{\pi/2}_{-\pi/2}\cos^4\theta \cdot d\theta $$

$$2\int \cos^4\theta \cdot d\theta = \frac{1}{2}\cos^3\theta\cdot \sin\theta+\frac{3}{2}\int \cos^2\theta d\theta $$$$= \frac{1}{2}\left[\cos^3\theta\cdot\sin\theta+\frac{3}{2}(\cos\theta \sin\theta+\theta)\right]$$

For ${-\pi/2}\le \theta \le {\pi/2}$, this will give a final answer of $\frac{3\pi}{4}$.