Question

If the volume of the solid in \( \mathbb{R}^3 \) bounded by the surfaces \[ x = -1, \ x = 1, \ y = -1, \ y = 1, \ z = 2, \ y^2 + z^2 = 2 \] is \( \alpha - \pi \), then \( \alpha = \text{.........}. \)

Hint:

For volume integrals involving circular boundaries, always use polar or cylindrical coordinates for easier computation.

Answer 1

Correct Answer: 5.99 - 6.01

Step 1: Visualize the region

The cylinder $y^2 + z^2 = 2$ has radius $\sqrt{2}$ centered on the x-axis.

The region is bounded:

• In x-direction: from -1 to 1
• In y-direction: from -1 to 1
• In z-direction: from the cylinder $z = \sqrt{2 - y^2}$ up to $z = 2$

Step 2: Check if the cylinder intersects the bounded region

At $y = 0$: $z^2 = 2 \Rightarrow z = \pm\sqrt{2} \approx \pm 1.414$

Since $\sqrt{2} < 2$ and the region extends from $y = -1$ to $y = 1$, we need to find where the cylinder intersects $y = \pm 1$:

At $y = \pm 1$: $1 + z^2 = 2 \Rightarrow z^2 = 1 \Rightarrow z = \pm 1$

Step 3: Set up the volume integral

The solid is bounded below by the cylinder and above by $z = 2$.

For $-1 \leq y \leq 1$, the cylinder gives $z = \sqrt{2 - y^2}$ (upper part).

$$V = \int_{-1}^{1} \int_{-1}^{1} \left(2 - \sqrt{2 - y^2}\right) dy, dx$$

Since the integral is independent of $x$:

$$V = 2 \int_{-1}^{1} \left(2 - \sqrt{2 - y^2}\right) dy$$

Step 4: Evaluate the integral

$$V = 2\left[\int_{-1}^{1} 2, dy - \int_{-1}^{1} \sqrt{2 - y^2}, dy\right]$$

First integral: $\int_{-1}^{1} 2, dy = 4$

Second integral: $\int_{-1}^{1} \sqrt{2 - y^2}, dy$

This represents the area under a semicircle of radius $\sqrt{2}$ from $y = -1$ to $y = 1$.

Using substitution $y = \sqrt{2}\sin\theta$:

$$\int_{-1}^{1} \sqrt{2 - y^2}, dy = \frac{\pi \cdot 2}{2} \cdot \frac{2}{\pi} + 2 \cdot \frac{1 \cdot 1}{2}$$

Using the formula for circular segment area:

Area $= r^2\arcsin(a/r) + a\sqrt{r^2 - a^2}$ where $r = \sqrt{2}$, $a = 1$

$$= 2\arcsin(1/\sqrt{2}) + 1\sqrt{2-1} = 2 \cdot \frac{\pi}{4} + 1 = \frac{\pi}{2} + 1$$

$$V = 2\left[4 - \left(\frac{\pi}{2} + 1\right)\right] = 2\left[3 - \frac{\pi}{2}\right] = 6 - \pi$$

Therefore: $\alpha - \pi = 6 - \pi$, so $\alpha = 6$

Answer: 6.00