It looks like you uploaded an image showing a region under the curve, and you want to find the volume of the solid obtained by rotating this region around the $y$-axis using the method of cylindrical shells.

Given the setup, here’s how to proceed with finding the volume:

Step 1: Identify the Curve and the Limits of Integration

From the image, the curve is given as: $y = 2x(1 - x)$ This curve is bounded between $x = 0$ and $x = 1$.

Step 2: Set Up the Cylindrical Shells

When rotating around the $y$-axis, we consider vertical strips.

• Radius ($r$): The radius of the shell is the distance from the y-axis to the strip at position $x$, which is simply $x$.

  $r = x$

• Height ($h$): The height of the shell is the value of the function $y = 2x(1-x)$.

  $h = 2x(1 - x)$

Step 3: Write the Volume Integral

The volume $V$ of the solid formed by rotating the region about the $y$-axis is given by:

$V = 2\pi \int_{x=0}^{x=1} r \cdot h \, dx$

Substituting $r = x$ and $h = 2x(1 - x)$:

$V = 2\pi \int_{0}^{1} x \cdot [2x(1 - x)] \, dx$

Step 4: Simplify and Evaluate the Integral

First, simplify the integrand:

$V = 2\pi \int_{0}^{1} 2x^2(1 - x) \, dx$

Expanding the integrand:

$V = 2\pi \int_{0}^{1} (2x^2 - 2x^3) \, dx$

Now, integrate each term:

$V = 2\pi \left[ \frac{2x^3}{3} - \frac{2x^4}{4} \right]_{0}^{1}$

Substituting the limits:

[ V = 2\pi \left[ \frac{2(1)^3}{3} - \frac{2(1)^4}{4} \right] - \left[ \frac{2(0)^3}{3} - \frac{2(0)^4}{4} \right] ]

This simplifies to:

$V = 2\pi \left[ \frac{2}{3} - \frac{1}{2} \right]$

Finding a common denominator and calculating:

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

Final Answer:

The volume of the solid obtained by rotating the region under the curve $y = 2x(1 - x)$ about the $y$-axis is:

$V = \frac{\pi}{3}$

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Do you have any questions or need further details?

Here are five related questions to consider:

1. What would the volume be if the region were rotated around the $x$-axis instead?
2. How does changing the limits of integration affect the volume of the solid?
3. What is the volume of the solid if the curve were $y = x^2$ instead?
4. Can you use the disk/washer method to find the volume and compare it with the shell method?
5. How would the solution change if the curve were shifted vertically or horizontally?

Tip: When setting up integrals for finding volumes, always sketch the region and the axis of rotation. This will help you visualize the problem better and avoid mistakes in setting up the integrand.