Sketch the region bounded by the curves y=7x−−√ and y=7x3
 then set up the integral needed to find the volume of the solid generated by revolving this region about the x
-axis using the shell method.

Let's solve this step by step:

### Step 1: Sketch the region bounded by the curves.

The given curves are:
1. \( y = \sqrt{7x} \)
2. \( y = 7x^3 \)

To understand the shape of the region, we first need to find where these curves intersect by setting \( \sqrt{7x} = 7x^3 \).

Solve this equation:
\[
\sqrt{7x} = 7x^3
\]
Square both sides:
\[
7x = 49x^6
\]
Simplify:
\[
7x - 49x^6 = 0
\]
Factor:
\[
7x(1 - 7x^5) = 0
\]
Thus, \( x = 0 \) or \( x = \frac{1}{\sqrt[5]{7}} \).

Therefore, the region of interest lies between \( x = 0 \) and \( x = \frac{1}{\sqrt[5]{7}} \).

### Step 2: Set up the integral for the volume using the shell method.

The shell method for revolving around the \( x \)-axis involves cylindrical shells with height given by the vertical distance between the two curves.

The height of the shell at a given \( x \) is the difference in \( y \)-values of the two curves:
\[
\text{Height of the shell} = \left( \sqrt{7x} - 7x^3 \right)
\]

The formula for the volume using the shell method is:
\[
V = 2\pi \int_{a}^{b} ( \text{radius} ) ( \text{height} ) \, dx
\]
For this problem:
- The **radius** is the distance from the axis of revolution (the \( x \)-axis), which is \( x \).
- The **height** is \( \sqrt{7x} - 7x^3 \).

Thus, the integral for the volume becomes:
\[
V = 2\pi \int_{0}^{\frac{1}{\sqrt[5]{7}}} x \left( \sqrt{7x} - 7x^3 \right) \, dx
\]

### Step 3: Simplify the integrand.

We can now distribute \( x \) to simplify the integrand:
\[
V = 2\pi \int_{0}^{\frac{1}{\sqrt[5]{7}}} \left( x \sqrt{7x} - 7x^4 \right) \, dx
\]

Now, simplify each term:
\[
x \sqrt{7x} = x (7x)^{1/2} = \sqrt{7} x^{3/2}
\]
So the volume integral becomes:
\[
V = 2\pi \int_{0}^{\frac{1}{\sqrt[5]{7}}} \left( \sqrt{7} x^{3/2} - 7x^4 \right) \, dx
\]

### Step 4: Final integral setup.

The integral to find the volume is:
\[
V = 2\pi \int_{0}^{\frac{1}{\sqrt[5]{7}}} \left( \sqrt{7} x^{3/2} - 7x^4 \right) \, dx
\]

This is the setup for the volume. You can now proceed to solve the integral if necessary.

---

Would you like me to evaluate the integral or go into further details? Let me know if you have any questions!

### 5 Related Questions:
1. How do we solve definite integrals involving fractional powers?
2. What is the difference between the disk/washer method and the shell method in volume problems?
3. How can we find the intersection points of two curves graphically?
4. How do we apply the shell method to revolve regions around the \( y \)-axis?
5. How can you visualize three-dimensional solids of revolution?

### Tip:
When setting up integrals for volumes, always ensure you correctly identify the radius and height of the shell or disk, based on the axis of revolution.

Sketch the region bounded by the curves y=√7x and y=7x^3 then set up the integral needed to find the volume of the solid generated by revolving this region about the x-axis using the shell method.

Learn how to find the volume of a solid generated by revolving the region bounded by y = √7x and y = 7x^3 around the x-axis using the shell method. This problem involves key calculus concepts such as curve sketching and the cylindrical shells formula.

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