Question 1166024
That's great that you've already solved the volume about the x-axis! Let's complete the remaining volume calculations using the disk/washer method for rotations about the $y$-axis and the vertical lines $x=3$ and $x=9$.

The region $R$ is bounded by:
* $y = \sqrt{x}$ (or $x = y^2$)
* $y = 0$ (the $x$-axis)
* $x = 3$
* $x = 9$

The region exists for $x$ from 3 to 9. The corresponding $y$ values range from $y = \sqrt{3}$ to $y = \sqrt{9} = 3$.

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## 1. Volume about the $x$-axis

You stated you found this to be $9\pi/2$. Let's confirm using the **Disk Method** with a vertical element (integration with respect to $x$):

$$V_x = \pi \int_{3}^{9} [f(x)]^2 dx$$
$$V_x = \pi \int_{3}^{9} (\sqrt{x})^2 dx$$
$$V_x = \pi \int_{3}^{9} x dx$$
$$V_x = \pi \left[ \frac{x^2}{2} \right]_{3}^{9}$$
$$V_x = \pi \left( \frac{9^2}{2} - \frac{3^2}{2} \right) = \pi \left( \frac{81}{2} - \frac{9}{2} \right) = \pi \left( \frac{72}{2} \right) = \mathbf{36\pi}$$

***Wait, your result of $9\pi/2$ seems incorrect. Please double-check your initial calculation. The correct volume is $36\pi$.***

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## 2. Volume about the $y$-axis

Since the axis of revolution ($x=0$) is a vertical line, we use the **Washer Method** with a horizontal element (integration with respect to $y$). The boundaries for $y$ are $\sqrt{3}$ and $3$.

The radii are measured from the $y$-axis ($x=0$) to the vertical boundaries $x=3$ and $x=9$:

* **Outer Radius ($R$):** The distance from $x=0$ to the farthest boundary $x=9$.
    $$R = 9 - 0 = 9$$
* **Inner Radius ($r$):** The distance from $x=0$ to the closest boundary $x = y^2$ (the curve).
    $$r = y^2 - 0 = y^2$$

$$V_y = \pi \int_{\sqrt{3}}^{3} \left(R^2 - r^2\right) dy$$
$$V_y = \pi \int_{\sqrt{3}}^{3} \left(9^2 - (y^2)^2\right) dy$$
$$V_y = \pi \int_{\sqrt{3}}^{3} (81 - y^4) dy$$
$$V_y = \pi \left[ 81y - \frac{y^5}{5} \right]_{\sqrt{3}}^{3}$$

$$V_y = \pi \left[ \left( 81(3) - \frac{3^5}{5} \right) - \left( 81(\sqrt{3}) - \frac{(\sqrt{3})^5}{5} \right) \right]$$
$$V_y = \pi \left[ \left( 243 - \frac{243}{5} \right) - \left( 81\sqrt{3} - \frac{9\sqrt{3}}{5} \right) \right]$$
$$V_y = \pi \left[ \left( \frac{1215 - 243}{5} \right) - \left( \frac{405\sqrt{3} - 9\sqrt{3}}{5} \right) \right]$$
$$V_y = \pi \left[ \frac{972}{5} - \frac{396\sqrt{3}}{5} \right]$$
$$V_y = \mathbf{\frac{972\pi}{5} - \frac{396\pi\sqrt{3}}{5} \text{ units}^3}$$ 

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## 3. Volume about the line $x=3$

Since the axis of revolution ($x=3$) is a vertical line, we integrate with respect to $y$ using the **Washer Method**.

The radii are measured from $x=3$ to the outer boundary $x=9$ and the inner boundary $x=y^2$:

* **Outer Radius ($R$):** The distance from $x=3$ to $x=9$.
    $$R = 9 - 3 = 6$$
* **Inner Radius ($r$):** The distance from $x=3$ to the curve $x = y^2$.
    $$r = y^2 - 3$$
    *Note: The integration limits remain $y=\sqrt{3}$ to $y=3$.*

$$V_{x=3} = \pi \int_{\sqrt{3}}^{3} \left(R^2 - r^2\right) dy$$
$$V_{x=3} = \pi \int_{\sqrt{3}}^{3} \left(6^2 - (y^2 - 3)^2\right) dy$$
$$V_{x=3} = \pi \int_{\sqrt{3}}^{3} \left(36 - (y^4 - 6y^2 + 9)\right) dy$$
$$V_{x=3} = \pi \int_{\sqrt{3}}^{3} (-y^4 + 6y^2 + 27) dy$$
$$V_{x=3} = \pi \left[ -\frac{y^5}{5} + \frac{6y^3}{3} + 27y \right]_{\sqrt{3}}^{3} = \pi \left[ -\frac{y^5}{5} + 2y^3 + 27y \right]_{\sqrt{3}}^{3}$$

$$V_{x=3} = \pi \left[ \left( -\frac{3^5}{5} + 2(3^3) + 27(3) \right) - \left( -\frac{(\sqrt{3})^5}{5} + 2(\sqrt{3})^3 + 27\sqrt{3} \right) \right]$$
$$V_{x=3} = \pi \left[ \left( -\frac{243}{5} + 54 + 81 \right) - \left( -\frac{9\sqrt{3}}{5} + 6\sqrt{3} + 27\sqrt{3} \right) \right]$$
$$V_{x=3} = \pi \left[ \left( 135 - \frac{243}{5} \right) - \left( 33\sqrt{3} - \frac{9\sqrt{3}}{5} \right) \right]$$
$$V_{x=3} = \pi \left[ \left( \frac{675 - 243}{5} \right) - \left( \frac{165\sqrt{3} - 9\sqrt{3}}{5} \right) \right]$$
$$V_{x=3} = \pi \left[ \frac{432}{5} - \frac{156\sqrt{3}}{5} \right]$$
$$V_{x=3} = \mathbf{\frac{432\pi}{5} - \frac{156\pi\sqrt{3}}{5} \text{ units}^3}$$

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## 4. Volume about the line $x=9$

Since the axis of revolution ($x=9$) is a vertical line, we integrate with respect to $y$ using the **Washer Method**.

The radii are measured from $x=9$ to the inner boundary $x=3$ and the outer boundary $x=y^2$:

* **Outer Radius ($R$):** The distance from $x=9$ to the curve $x=y^2$.
    $$R = 9 - y^2$$
* **Inner Radius ($r$):** The distance from $x=9$ to $x=3$.
    $$r = 9 - 3 = 6$$
    *Note: The integration limits remain $y=\sqrt{3}$ to $y=3$.*

$$V_{x=9} = \pi \int_{\sqrt{3}}^{3} \left(R^2 - r^2\right) dy$$
$$V_{x=9} = \pi \int_{\sqrt{3}}^{3} \left((9 - y^2)^2 - 6^2\right) dy$$
$$V_{x=9} = \pi \int_{\sqrt{3}}^{3} \left((81 - 18y^2 + y^4) - 36\right) dy$$
$$V_{x=9} = \pi \int_{\sqrt{3}}^{3} (y^4 - 18y^2 + 45) dy$$
$$V_{x=9} = \pi \left[ \frac{y^5}{5} - \frac{18y^3}{3} + 45y \right]_{\sqrt{3}}^{3} = \pi \left[ \frac{y^5}{5} - 6y^3 + 45y \right]_{\sqrt{3}}^{3}$$

$$V_{x=9} = \pi \left[ \left( \frac{3^5}{5} - 6(3^3) + 45(3) \right) - \left( \frac{(\sqrt{3})^5}{5} - 6(\sqrt{3})^3 + 45\sqrt{3} \right) \right]$$
$$V_{x=9} = \pi \left[ \left( \frac{243}{5} - 162 + 135 \right) - \left( \frac{9\sqrt{3}}{5} - 18\sqrt{3} + 45\sqrt{3} \right) \right]$$
$$V_{x=9} = \pi \left[ \left( \frac{243}{5} - 27 \right) - \left( 27\sqrt{3} + \frac{9\sqrt{3}}{5} \right) \right]$$
$$V_{x=9} = \pi \left[ \left( \frac{243 - 135}{5} \right) - \left( \frac{135\sqrt{3} + 9\sqrt{3}}{5} \right) \right]$$
$$V_{x=9} = \pi \left[ \frac{108}{5} - \frac{144\sqrt{3}}{5} \right]$$
$$V_{x=9} = \mathbf{\frac{108\pi}{5} - \frac{144\pi\sqrt{3}}{5} \text{ units}^3}$$