ChatGPT vs Grok for Math

We want the volume of the solid formed by rotating the region bounded by:

• $y = x^2$
• $y = 0$
• $x = 2$

about the $y$-axis.

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1) Understand the region

The curve $y = x^2$ is a parabola opening upward.

The line $y=0$ is the $x$-axis.

The line $x=2$ is a vertical line.

So the bounded region is the area under $y=x^2$, above the $x$-axis, from $x=0$ to $x=2$.

Why from $x=0$? Because $y=x^2$ meets $y=0$ at:

$$x^2=0 \Rightarrow x=0$$

So the region runs from $x=0$ to $x=2$.

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2) Choose a method

We are rotating around the $y$-axis.

Since the region is naturally described in terms of $x$, the shell method is the easiest.

Shell method formula

$$V = 2\pi \int_a^b (\text{radius})(\text{height})\,dx$$

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3) Set up the shell

Take a vertical strip at position $x$.

When rotated about the $y$-axis:

• radius = distance from $x$ to the $y$-axis = $x$
• height = top minus bottom = $x^2 - 0 = x^2$

So:

$$V = 2\pi \int_0^2 x(x^2)\,dx$$

Simplify:

$$V = 2\pi \int_0^2 x^3\,dx$$

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4) Evaluate the integral

$$\int x^3\,dx = \frac{x^4}{4}$$

So:

$$V = 2\pi \left[\frac{x^4}{4}\right]_0^2$$

Substitute the bounds:

$$V = 2\pi \left(\frac{2^4}{4} - \frac{0^4}{4}\right)$$

$$V = 2\pi \left(\frac{16}{4}\right)$$

$$V = 2\pi(4)$$

$$V = 8\pi$$

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5) Final answer

$$\boxed{8\pi}$$

So the volume of the solid is:

$$\boxed{8\pi \text{ cubic units}}$$

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Quick reasoning summary

• The region goes from $x=0$ to $x=2$.
• Rotating about the $y$-axis suggests using cylindrical shells.
• Each shell has radius $x$ and height $x^2$.
• Volume: $$V=2\pi\int_0^2 x(x^2)\,dx=2\pi\int_0^2 x^3\,dx=8\pi$$

If you want, I can also solve it using the washer method for comparison.

Step 1: Understand the Region and Setup

The region is bounded by $y = x^2$ (upper boundary), $y = 0$ (x-axis, lower boundary), and $x = 2$ (right boundary). It implicitly starts at $x = 0$ (left boundary, where $y = x^2$ intersects $y = 0$). Thus, the region lies between $x = 0$ and $x = 2$, with $0 \leq y \leq x^2$.

We rotate this region about the y-axis to form a solid of revolution. Two common methods are cylindrical shells (integrating with respect to $x$) or washers/disks (integrating with respect to $y$). Both yield the same result; we'll use shells first (simpler here, as the region is described naturally in $x$), then verify with washers.

Step 2: Cylindrical Shells Method

• A vertical strip at position $x$ (where $0 \leq x \leq 2$) has:
  • Height: $x^2 - 0 = x^2$.
  • Radius: Distance from the y-axis, which is $x$.
  • Thickness: $dx$.
• Rotating this strip forms a cylindrical shell with surface area $2\pi \cdot (\text{radius}) \cdot (\text{height})$ and volume element $2\pi x \cdot x^2 \, dx = 2\pi x^3 \, dx$.
• Integrate from $x = 0$ to $x = 2$: $$V = \int_0^2 2\pi x^3 \, dx = 2\pi \int_0^2 x^3 \, dx = 2\pi \left[ \frac{x^4}{4} \right]_0^2 = 2\pi \left( \frac{2^4}{4} - 0 \right) = 2\pi \cdot \frac{16}{4} = 2\pi \cdot 4 = 8\pi.$$

Step 3: Verification Using Washer Method

To confirm, rewrite in terms of $y$:

• $y = x^2$ implies $x = \sqrt{y}$ (since $x \geq 0$).
• $y$ ranges from 0 to $4$ (maximum at $x=2$, $y=4$).
• For fixed $y$ ($0 \leq y \leq 4$), the horizontal strip extends from $x = \sqrt{y}$ (left, on the parabola) to $x = 2$ (right boundary).
• Rotation about the y-axis forms a washer:
  • Outer radius: $2$.
  • Inner radius: $\sqrt{y}$.
• Volume element: $\pi (2^2 - (\sqrt{y})^2) \, dy = \pi (4 - y) \, dy$.
• Integrate from $y = 0$ to $y = 4$: $$V = \int_0^4 \pi (4 - y) \, dy = \pi \left[ 4y - \frac{y^2}{2} \right]_0^4 = \pi \left( (4 \cdot 4 - \frac{16}{2}) - 0 \right) = \pi (16 - 8) = 8\pi.$$

Both methods agree, confirming the volume.

Final Answer

The volume is $8\pi$.