Volume of Revolution About the Y-Axis

Setting Up the Problem

Visualizing the region: We have a parabola y = x², the x-axis (y = 0), and the vertical line x = 2. This creates a bounded region in the first quadrant.

Choosing the method: Since we're rotating about the y-axis and our region is naturally described in terms of x, I'll use the Shell Method — it's elegantly suited here.

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The Shell Method Formula

$$V = 2\pi \int_a^b \underbrace{x}_{\text{radius}} \cdot \underbrace{f(x)}_{\text{height}} \, dx$$

Physical intuition: Imagine thin cylindrical shells, each at distance x from the y-axis, with height equal to the function value at that x.

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Identifying the Components

Component	Value
Radius of shell	x
Height of shell	x² - 0 = x²
Limits of integration	x = 0 to x = 2

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Computing the Integral

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

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

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

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

$$V = 2\pi \left(\frac{16}{4} - 0\right)$$

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

$$\boxed{V = 8\pi \approx 25.13 \text{ cubic units}}$$

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Verification Using the Disk/Washer Method

Rotating about the y-axis means integrating with respect to y, from y = 0 to y = 4.

From y = x², we get x = √y (the outer radius), and x = 2 creates a washer.

$$V = \pi \int_0^4 \left[(\underbrace{2}_{\text{outer}})^2 - (\underbrace{\sqrt{y}}_{\text{inner}})^2\right] dy$$

$$V = \pi \int_0^4 (4 - y) \, dy = \pi \left[4y - \frac{y^2}{2}\right]_0^4$$

$$V = \pi \left(16 - \frac{16}{2}\right) = \pi(16 - 8) = 8\pi \checkmark$$

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Summary

Both methods confirm:

$$V = 8\pi \text{ cubic units}$$

The Shell Method was more direct (one clean integral in x), while the Washer Method required rewriting x as a function of y — but both are powerful tools depending on your geometry.

Alright, let’s solve it step-by-step.

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Step 1 – Understand the region

We are given:

• Curve: $y = x^2$
• Horizontal line: $y = 0$ (the x-axis)
• Vertical line: $x = 2$

Boundaries: From $y = 0$ to $y = x^2$, with $x$ going from $0$ to $2$.

In the $xy$-plane: The region is bounded above by $y = x^2$, below by $y = 0$, right by $x = 2$, left by $x = 0$ (since $y = x^2$ starts at origin).

Actually, more precisely: The region is between $y = 0$ and $y = x^2$, for $0 \le x \le 2$. That means the top boundary is $y = x^2$ and bottom is $y = 0$.

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Step 2 – Revolving around the y-axis

We revolve this region around the y-axis.

Sketch: The curve $y = x^2$ from $x = 0$ to $x = 2$, and the line $x = 2$, below it $y = 0$.

When revolved around the y-axis, the solid has two parts:

1. For a given $y$ from $y = 0$ to $y = 4$ ($y = 2^2$), we have:

   • The inner radius: from y-axis to $x = \sqrt{y}$ (since $y = x^2$ means $x = \sqrt{y}$ in first quadrant).
   • The outer radius: $x = 2$ (from the line $x = 2$) for $y$ from 0 to 4.
   • Wait: the region in the $x$-direction goes from $x = \sqrt{y}$ to $x = 2$ only for $y$ such that $\sqrt{y} \le 2$ i.e. $y \le 4$.
     That’s exactly the region for $0 \le y \le 4$.

   But from the description: “region bounded by $y = x^2$, $y = 0$, and $x = 2$” means all points with $0 \le y \le x^2$ and $0 \le x \le 2$.

   This is vertical slice description:

   For a fixed $x \in [0, 2]$, $y$ ranges from 0 to $x^2$.

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Step 3 – Choose method: Washer or Shell?

Because we’re revolving around the y-axis, two options:

(1) Washer method (integrating with respect to $y$):

For $0 \le y \le 4$, cross-section perpendicular to y-axis:
Radii:
Outer radius = $R(y) = 2$ (from y-axis to line $x = 2$)
Inner radius = $r(y) = \sqrt{y}$ (from y-axis to curve $x = \sqrt{y}$).

But careful: This is only valid if the region in the y-slice is between $x = \sqrt{y}$ and $x = 2$, for $0 \le y \le 4$.
Wait—is that correct? Let’s visualize the region: In the xy-plane, the region is under $y = x^2$ from $x = 0$ to $x = 2$.

If we take a horizontal line at $y$, it hits the region at $x$ from $\sqrt{y}$ to $2$? No—that’s wrong! That would be the part to the right of the curve $x = \sqrt{y}$. But our region is to the left of the curve $y = x^2$, i.e. $0 \le y \le x^2$.

So for a given $x$, $0 \le y \le x^2$. For a given $y$, what $x$ satisfy $y \le x^2$ and $x \le 2$?
$x^2 \ge y$ ⇒ $x \ge \sqrt{y}$ (since $x \ge 0$). So for a fixed $y$, $x$ ranges from $\sqrt{y}$ to 2.

Yes! So I was correct the first time: The horizontal slice at height $y$ from 0 to 4 goes from $x = \sqrt{y}$ to $x = 2$.

So indeed, region is { $\sqrt{y} \le x \le 2, 0 \le y \le 4$}.

Now revolving around y-axis:
Washer’s outer radius = $R = 2$ (distance from y-axis to $x = 2$)
Washer’s inner radius = $r = \sqrt{y}$ (distance from y-axis to $x = \sqrt{y}$)
Area of washer = $\pi [R^2 - r^2] = \pi [4 - y]$.

So volume =
$$V = \int_{y=0}^{4} \pi [4 - y] \, dy$$

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(2) Check with shell method too (for consistency):

Shell method (integrating with respect to $x$):

For $0 \le x \le 2$, take vertical strip at $x$ of height $x^2$, revolved around y-axis ⇒ cylindrical shell radius = $x$, height = $x^2$, thickness = $dx$.

Shell surface area = $2\pi (\text{radius})(\text{height}) = 2\pi x \cdot x^2 = 2\pi x^3$.

So
$$V = \int_{x=0}^{2} 2\pi x^3 \, dx$$

Let’s compute both to see if they match.

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Step 4 – Compute via washer method

$$V = \pi \int_{0}^{4} (4 - y) \, dy$$ $$= \pi \left[ 4y - \frac{y^2}{2} \right]_{0}^{4}$$ $$= \pi \left[ (16) - \frac{16}{2} \right]$$ $$= \pi [16 - 8]$$ $$= 8\pi$$

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Step 5 – Compute via shell method

$$V = 2\pi \int_{0}^{2} x^3 \, dx$$ $$= 2\pi \left[ \frac{x^4}{4} \right]_{0}^{2}$$ $$= 2\pi \cdot \frac{16}{4}$$ $$= 2\pi \cdot 4 = 8\pi$$

Both match.

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Step 6 – Conclusion

The volume of the solid is $8\pi$ cubic units.

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$$\boxed{8\pi}$$