To find the volume of the solid generated by rotating the given region about the y-axis using the method of cylindrical shells, we follow these steps:

Step 1: Identify the Shell Elements

The volume of a solid of revolution using the cylindrical shells method is given by:

$V = \int_{a}^{b} 2\pi ( \text{radius} ) ( \text{height} ) \, dx$

• Radius of shell: Since we rotate about the y-axis, the radius of a shell at any point $x$ is simply $r = x$.
• Height of shell: The height is given by the function $y = \cos\left(\frac{\pi x}{2}\right)$.
• Limits of integration: The region is bounded by $x = 0$ and $x = 1$, so we integrate from $x = 0$ to $x = 1$.

Step 2: Set Up the Integral

$V = \int_0^1 2\pi x \cos\left(\frac{\pi x}{2}\right) \, dx$

Step 3: Solve the Integral

We solve:

$I = \int x \cos\left(\frac{\pi x}{2}\right) dx$

Using integration by parts, let:

• $u = x \Rightarrow du = dx$
• $dv = \cos\left(\frac{\pi x}{2}\right) dx$

We integrate $dv$ using substitution: Let $w = \frac{\pi x}{2} \Rightarrow dw = \frac{\pi}{2} dx$ so that:

$dx = \frac{2}{\pi} dw$

Thus,

$\int \cos(w) dx = \int \cos(w) \frac{2}{\pi} dw = \frac{2}{\pi} \sin(w) = \frac{2}{\pi} \sin\left(\frac{\pi x}{2}\right)$

Now applying integration by parts:

$I = x \cdot \frac{2}{\pi} \sin\left(\frac{\pi x}{2}\right) - \int \frac{2}{\pi} \sin\left(\frac{\pi x}{2}\right) dx$

Since we already found:

$\int \cos\left(\frac{\pi x}{2}\right) dx = \frac{2}{\pi} \sin\left(\frac{\pi x}{2}\right)$

It follows that:

$I = x \cdot \frac{2}{\pi} \sin\left(\frac{\pi x}{2}\right) - \frac{4}{\pi^2} \cos\left(\frac{\pi x}{2}\right)$

Evaluating from $x = 0$ to $x = 1$:

At $x = 1$:

$I(1) = 1 \cdot \frac{2}{\pi} \sin\left(\frac{\pi}{2}\right) - \frac{4}{\pi^2} \cos\left(\frac{\pi}{2}\right)$

Since $\sin(\pi/2) = 1$ and $\cos(\pi/2) = 0$:

$I(1) = \frac{2}{\pi}$

At $x = 0$:

$I(0) = 0 \cdot \frac{2}{\pi} \sin(0) - \frac{4}{\pi^2} \cos(0) = -\frac{4}{\pi^2}$

Thus,

$I(1) - I(0) = \frac{2}{\pi} - \left(-\frac{4}{\pi^2}\right) = \frac{2}{\pi} + \frac{4}{\pi^2}$

Step 4: Compute the Final Volume

$V = 2\pi \left( \frac{2}{\pi} + \frac{4}{\pi^2} \right)$

$V = 4 + \frac{8}{\pi}$

Final Answer:

$V = 4 + \frac{8}{\pi}$

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Related Problems:

1. Compute the volume if the region is rotated about $x = -1$ instead of the $y$-axis.
2. Find the volume of the same region using the disk/washer method.
3. Determine the surface area of the solid obtained after rotation.
4. Generalize the result by changing the function to $y = \cos(kx)$.
5. Compute the centroid of the original bounded region.

Tip:

In the cylindrical shells method, always check the radius and height carefully before setting up the integral!