Using your calculator, determine the surface area obtained when the curve y = x^3 + x is revolved around the x-axis over the interval of x  = [1, 3]. SHow the integral you used.


To determine the surface area obtained when the curve \( y = x^3 + x \) is revolved around the x-axis over the interval \( x = [1, 3] \), we use the formula for the surface area of a solid of revolution around the x-axis:

\[
A = 2\pi \int_{a}^{b} y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
\]

### Step 1: Differentiate \( y = x^3 + x \)

The first step is to find the derivative \( \frac{dy}{dx} \).

\[
\frac{dy}{dx} = 3x^2 + 1
\]

### Step 2: Set up the integral

Now, substitute \( y = x^3 + x \) and \( \frac{dy}{dx} = 3x^2 + 1 \) into the surface area formula:

\[
A = 2\pi \int_{1}^{3} (x^3 + x) \sqrt{1 + (3x^2 + 1)^2} \, dx
\]

### Step 3: Simplify the expression under the square root

Expand \( (3x^2 + 1)^2 \):

\[
(3x^2 + 1)^2 = 9x^4 + 6x^2 + 1
\]

So the integral becomes:

\[
A = 2\pi \int_{1}^{3} (x^3 + x) \sqrt{1 + 9x^4 + 6x^2 + 1} \, dx
\]

Simplify the expression under the square root:

\[
A = 2\pi \int_{1}^{3} (x^3 + x) \sqrt{9x^4 + 6x^2 + 2} \, dx
\]

### Step 4: Numerical Integration

Now we compute this integral numerically.

Let me calculate the value.The surface area obtained when the curve \( y = x^3 + x \) is revolved around the x-axis over the interval \( x = [1, 3] \) is approximately \( 2819.72 \) square units.

The integral used to compute the surface area was:

\[
A = 2\pi \int_{1}^{3} (x^3 + x) \sqrt{9x^4 + 6x^2 + 2} \, dx
\]

Would you like further details on any part of this solution, or do you have any other questions?

Here are five related questions you could explore:
1. What happens to the surface area if the curve \( y = x^3 - x \) is revolved around the x-axis?
2. How would the surface area change if we revolve the curve around the y-axis instead?
3. What is the volume of the solid formed when this same curve is revolved around the x-axis?
4. How can we compute the surface area for parametric curves?
5. What role does the derivative \( \frac{dy}{dx} \) play in the surface area formula?

**Tip:** Numerical integration can often be used when dealing with complex expressions, especially when the integrand cannot be simplified easily for analytical solutions.

Using your calculator, determine the surface area obtained when the curve y = x^3 + x is revolved around the x-axis over the interval of x = [1, 3]. Show the integral you used.

Calculate the surface area when the curve y = x^3 + x is revolved around the x-axis over the interval x = [1, 3]. This problem involves calculus concepts, including surface area of revolution and numerical integration, and is suitable for undergraduate calculus students.

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