To find the surface area of the solid generated by rotating the curve $y = 4x^2$ around the y-axis, we use the formula for the surface area of a solid of revolution. The surface area $A$ is given by:

$A = 2 \pi \int_{x_1}^{x_2} \left( x \cdot \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \right) \, dx$

Step 1: Find the derivative of $y = 4x^2$

The curve is $y = 4x^2$. The derivative of $y$ with respect to $x$ is:

$\frac{dy}{dx} = 8x$

Step 2: Set up the surface area integral

We are interested in the surface area for $x$ between $-1$ and $1$. Therefore, the integral becomes:

$A = 2\pi \int_{-1}^{1} \left( x \cdot \sqrt{1 + (8x)^2} \right) \, dx$ $A = 2\pi \int_{-1}^{1} \left( x \cdot \sqrt{1 + 64x^2} \right) \, dx$

Step 3: Solve the integral

This integral is symmetric, so we can simplify by considering only the interval from $0$ to $1$, and then doubling the result:

$A = 4\pi \int_0^1 x \cdot \sqrt{1 + 64x^2} \, dx$

This integral can be solved using substitution. Let:

$u = 1 + 64x^2 \quad \text{so} \quad du = 128x \, dx$

Thus, the integral becomes:

$A = 4\pi \cdot \frac{1}{128} \int_1^{65} \sqrt{u} \, du$ $A = \frac{\pi}{32} \int_1^{65} \sqrt{u} \, du$

Step 4: Integrate

We can now integrate $\sqrt{u}$:

$\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}$

Evaluating from $1$ to $65$:

$A = \frac{\pi}{32} \left[ \frac{2}{3} u^{3/2} \right]_1^{65}$ $A = \frac{\pi}{32} \left( \frac{2}{3} \left( 65^{3/2} - 1^{3/2} \right) \right)$ $A = \frac{\pi}{48} \left( 65^{3/2} - 1 \right)$

Step 5: Approximate $65^{3/2}$

Now, let's approximate $65^{3/2}$. We can write:

$65^{3/2} = 65 \times \sqrt{65} \approx 65 \times 8.062 = 524.03$

So:

$A = \frac{\pi}{48} \left( 524.03 - 1 \right) = \frac{\pi}{48} \times 523.03$

Finally:

$A \approx \frac{\pi \times 523.03}{48} \approx 34.4 \, \text{square units}$

Thus, the surface area of the solid is approximately $34.4$ square units.

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Would you like further details or clarifications on any of the steps?

Here are 5 related questions:

1. What is the formula for the surface area of a solid of revolution?
2. How do you compute the surface area of a solid generated by rotating a curve around the x-axis?
3. What other methods can be used to solve integrals like this one?
4. How do you perform the substitution $u = 1 + 64x^2$ in the integral?
5. What is the geometric interpretation of a surface of revolution?

Tip: When dealing with symmetric curves like this one, always check if the integral is symmetric, as this can simplify your calculations by halving the range and doubling the result.