R is the region bounded above by the graph of f(x)=6e−x2 and below by the x-axis over the interval [1,2]. Find the volume of the solid of revolution formed by revolving R around the y-axis.

To find the volume of the solid formed by revolving the region \( R \) around the \( y \)-axis, we use the method known as the **shell method**. The volume is given by the integral formula:

\[
V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx
\]

Where:
- \( f(x) = 6e^{-x^2} \) is the function being revolved,
- \( a = 1 \) and \( b = 2 \) are the limits of integration (from the interval \([1, 2]\)).

Now, we can set up the volume integral:

\[
V = 2\pi \int_{1}^{2} x \cdot 6e^{-x^2} \, dx
\]

To solve this, we apply substitution:
Let \( u = x^2 \), then \( du = 2x \, dx \), or \( \frac{du}{2} = x \, dx \).

When \( x = 1 \), \( u = 1^2 = 1 \), and when \( x = 2 \), \( u = 2^2 = 4 \).

Thus, the integral becomes:

\[
V = 2\pi \cdot 6 \cdot \frac{1}{2} \int_{1}^{4} e^{-u} \, du
\]

Simplifying:

\[
V = 6\pi \int_{1}^{4} e^{-u} \, du
\]

The integral of \( e^{-u} \) is \( -e^{-u} \), so we evaluate:

\[
V = 6\pi \left[ -e^{-u} \right]_{1}^{4}
\]

Substitute the limits:

\[
V = 6\pi \left( -e^{-4} + e^{-1} \right)
\]

Simplifying further:

\[
V = 6\pi \left( e^{-1} - e^{-4} \right)
\]

Finally, this is the volume of the solid of revolution.

If you want a numerical value:

\[
V \approx 6\pi (0.3679 - 0.0183) \approx 6\pi \cdot 0.3496 \approx 6.59 \, \text{units}^3
\]

### Do you want a further breakdown of any of the steps? 

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Here are 5 related questions to consider:

1. How would the result change if the region is revolved around the \( x \)-axis?
2. Can you apply the washer method for this problem? How does it differ from the shell method?
3. What if the function were \( f(x) = e^{-x^2} \) instead of \( 6e^{-x^2} \)?
4. How would you approach the problem if the bounds were \( [0, 1] \) instead of \( [1, 2] \)?
5. What does the region \( R \) look like when plotted, and how does that affect the solid's shape?

**Tip:** When solving volume of revolution problems, always visualize the shape of the region and the axis of rotation to determine the most suitable method—washer, disk, or shell method.

R is the region bounded above by the graph of f(x)=6e−x^2 and below by the x-axis over the interval [1,2]. Find the volume of the solid of revolution formed by revolving R around the y-axis.

Explore the solution to finding the volume of the solid of revolution formed by revolving the region bounded by f(x) = 6e^(-x^2) and the x-axis over [1, 2] around the y-axis using the shell method. This step-by-step guide is perfect for college-level calculus students.

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