document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1166024>Question 1166024</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the volumes of the solids generated by revolving the regions bounded by the graphs about the given lines: y=sqrtx, y=0, x=3, and x=9 about the x=axis, the y-axis, the line at x=3, and the line at x=9. Use integration and washer/disk method. I have already solved about the x-axis to be 9/2 pi.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852872 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2138)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=852872\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 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 $x=9$.\r<BR>\n" );
document.write( "\n" );
document.write( "The region $R$ is bounded by:<BR>\n" );
document.write( "* $y = \sqrt{x}$ (or $x = y^2$)<BR>\n" );
document.write( "* $y = 0$ (the $x$-axis)<BR>\n" );
document.write( "* $x = 3$<BR>\n" );
document.write( "* $x = 9$\r<BR>\n" );
document.write( "\n" );
document.write( "The region exists for $x$ from 3 to 9. The corresponding $y$ values range from $y = \sqrt{3}$ to $y = \sqrt{9} = 3$.\r<BR>\n" );
document.write( "\n" );
document.write( "---\r<BR>\n" );
document.write( "\n" );
document.write( "## 1. Volume about the $x$-axis\r<BR>\n" );
document.write( "\n" );
document.write( "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 $x$):\r<BR>\n" );
document.write( "\n" );
document.write( "$$V_x = \pi \int_{3}^{9} [f(x)]^2 dx$$<BR>\n" );
document.write( "$$V_x = \pi \int_{3}^{9} (\sqrt{x})^2 dx$$<BR>\n" );
document.write( "$$V_x = \pi \int_{3}^{9} x dx$$<BR>\n" );
document.write( "$$V_x = \pi \left[ \frac{x^2}{2} \right]_{3}^{9}$$<BR>\n" );
document.write( "$$V_x = \pi \left( \frac{9^2}{2} - \frac{3^2}{2} \right) = \pi \left( \frac{81}{2} - \frac{9}{2} \right) = \pi \left( \frac{72}{2} \right) = \mathbf{36\pi}$$\r<BR>\n" );
document.write( "\n" );
document.write( "***Wait, your result of $9\pi/2$ seems incorrect. Please double-check your initial calculation. The correct volume is $36\pi$.***\r<BR>\n" );
document.write( "\n" );
document.write( "---\r<BR>\n" );
document.write( "\n" );
document.write( "## 2. Volume about the $y$-axis\r<BR>\n" );
document.write( "\n" );
document.write( "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 $3$.\r<BR>\n" );
document.write( "\n" );
document.write( "The radii are measured from the $y$-axis ($x=0$) to the vertical boundaries $x=3$ and $x=9$:\r<BR>\n" );
document.write( "\n" );
document.write( "* **Outer Radius ($R$):** The distance from $x=0$ to the farthest boundary $x=9$.<BR>\n" );
document.write( "    $$R = 9 - 0 = 9$$<BR>\n" );
document.write( "* **Inner Radius ($r$):** The distance from $x=0$ to the closest boundary $x = y^2$ (the curve).<BR>\n" );
document.write( "    $$r = y^2 - 0 = y^2$$\r<BR>\n" );
document.write( "\n" );
document.write( "$$V_y = \pi \int_{\sqrt{3}}^{3} \left(R^2 - r^2\right) dy$$<BR>\n" );
document.write( "$$V_y = \pi \int_{\sqrt{3}}^{3} \left(9^2 - (y^2)^2\right) dy$$<BR>\n" );
document.write( "$$V_y = \pi \int_{\sqrt{3}}^{3} (81 - y^4) dy$$<BR>\n" );
document.write( "$$V_y = \pi \left[ 81y - \frac{y^5}{5} \right]_{\sqrt{3}}^{3}$$\r<BR>\n" );
document.write( "\n" );
document.write( "$$V_y = \pi \left[ \left( 81(3) - \frac{3^5}{5} \right) - \left( 81(\sqrt{3}) - \frac{(\sqrt{3})^5}{5} \right) \right]$$<BR>\n" );
document.write( "$$V_y = \pi \left[ \left( 243 - \frac{243}{5} \right) - \left( 81\sqrt{3} - \frac{9\sqrt{3}}{5} \right) \right]$$<BR>\n" );
document.write( "$$V_y = \pi \left[ \left( \frac{1215 - 243}{5} \right) - \left( \frac{405\sqrt{3} - 9\sqrt{3}}{5} \right) \right]$$<BR>\n" );
document.write( "$$V_y = \pi \left[ \frac{972}{5} - \frac{396\sqrt{3}}{5} \right]$$<BR>\n" );
document.write( "$$V_y = \mathbf{\frac{972\pi}{5} - \frac{396\pi\sqrt{3}}{5} \text{ units}^3}$$ \r<BR>\n" );
document.write( "\n" );
document.write( "---\r<BR>\n" );
document.write( "\n" );
document.write( "## 3. Volume about the line $x=3$\r<BR>\n" );
document.write( "\n" );
document.write( "Since the axis of revolution ($x=3$) is a vertical line, we integrate with respect to $y$ using the **Washer Method**.\r<BR>\n" );
document.write( "\n" );
document.write( "The radii are measured from $x=3$ to the outer boundary $x=9$ and the inner boundary $x=y^2$:\r<BR>\n" );
document.write( "\n" );
document.write( "* **Outer Radius ($R$):** The distance from $x=3$ to $x=9$.<BR>\n" );
document.write( "    $$R = 9 - 3 = 6$$<BR>\n" );
document.write( "* **Inner Radius ($r$):** The distance from $x=3$ to the curve $x = y^2$.<BR>\n" );
document.write( "    $$r = y^2 - 3$$<BR>\n" );
document.write( "    *Note: The integration limits remain $y=\sqrt{3}$ to $y=3$.*\r<BR>\n" );
document.write( "\n" );
document.write( "$$V_{x=3} = \pi \int_{\sqrt{3}}^{3} \left(R^2 - r^2\right) dy$$<BR>\n" );
document.write( "$$V_{x=3} = \pi \int_{\sqrt{3}}^{3} \left(6^2 - (y^2 - 3)^2\right) dy$$<BR>\n" );
document.write( "$$V_{x=3} = \pi \int_{\sqrt{3}}^{3} \left(36 - (y^4 - 6y^2 + 9)\right) dy$$<BR>\n" );
document.write( "$$V_{x=3} = \pi \int_{\sqrt{3}}^{3} (-y^4 + 6y^2 + 27) dy$$<BR>\n" );
document.write( "$$V_{x=3} = \pi \left[ -\frac{y^5}{5} + \frac{6y^3}{3} + 27y \right]_{\sqrt{3}}^{3} = \pi \left[ -\frac{y^5}{5} + 2y^3 + 27y \right]_{\sqrt{3}}^{3}$$\r<BR>\n" );
document.write( "\n" );
document.write( "$$V_{x=3} = \pi \left[ \left( -\frac{3^5}{5} + 2(3^3) + 27(3) \right) - \left( -\frac{(\sqrt{3})^5}{5} + 2(\sqrt{3})^3 + 27\sqrt{3} \right) \right]$$<BR>\n" );
document.write( "$$V_{x=3} = \pi \left[ \left( -\frac{243}{5} + 54 + 81 \right) - \left( -\frac{9\sqrt{3}}{5} + 6\sqrt{3} + 27\sqrt{3} \right) \right]$$<BR>\n" );
document.write( "$$V_{x=3} = \pi \left[ \left( 135 - \frac{243}{5} \right) - \left( 33\sqrt{3} - \frac{9\sqrt{3}}{5} \right) \right]$$<BR>\n" );
document.write( "$$V_{x=3} = \pi \left[ \left( \frac{675 - 243}{5} \right) - \left( \frac{165\sqrt{3} - 9\sqrt{3}}{5} \right) \right]$$<BR>\n" );
document.write( "$$V_{x=3} = \pi \left[ \frac{432}{5} - \frac{156\sqrt{3}}{5} \right]$$<BR>\n" );
document.write( "$$V_{x=3} = \mathbf{\frac{432\pi}{5} - \frac{156\pi\sqrt{3}}{5} \text{ units}^3}$$\r<BR>\n" );
document.write( "\n" );
document.write( "---\r<BR>\n" );
document.write( "\n" );
document.write( "## 4. Volume about the line $x=9$\r<BR>\n" );
document.write( "\n" );
document.write( "Since the axis of revolution ($x=9$) is a vertical line, we integrate with respect to $y$ using the **Washer Method**.\r<BR>\n" );
document.write( "\n" );
document.write( "The radii are measured from $x=9$ to the inner boundary $x=3$ and the outer boundary $x=y^2$:\r<BR>\n" );
document.write( "\n" );
document.write( "* **Outer Radius ($R$):** The distance from $x=9$ to the curve $x=y^2$.<BR>\n" );
document.write( "    $$R = 9 - y^2$$<BR>\n" );
document.write( "* **Inner Radius ($r$):** The distance from $x=9$ to $x=3$.<BR>\n" );
document.write( "    $$r = 9 - 3 = 6$$<BR>\n" );
document.write( "    *Note: The integration limits remain $y=\sqrt{3}$ to $y=3$.*\r<BR>\n" );
document.write( "\n" );
document.write( "$$V_{x=9} = \pi \int_{\sqrt{3}}^{3} \left(R^2 - r^2\right) dy$$<BR>\n" );
document.write( "$$V_{x=9} = \pi \int_{\sqrt{3}}^{3} \left((9 - y^2)^2 - 6^2\right) dy$$<BR>\n" );
document.write( "$$V_{x=9} = \pi \int_{\sqrt{3}}^{3} \left((81 - 18y^2 + y^4) - 36\right) dy$$<BR>\n" );
document.write( "$$V_{x=9} = \pi \int_{\sqrt{3}}^{3} (y^4 - 18y^2 + 45) dy$$<BR>\n" );
document.write( "$$V_{x=9} = \pi \left[ \frac{y^5}{5} - \frac{18y^3}{3} + 45y \right]_{\sqrt{3}}^{3} = \pi \left[ \frac{y^5}{5} - 6y^3 + 45y \right]_{\sqrt{3}}^{3}$$\r<BR>\n" );
document.write( "\n" );
document.write( "$$V_{x=9} = \pi \left[ \left( \frac{3^5}{5} - 6(3^3) + 45(3) \right) - \left( \frac{(\sqrt{3})^5}{5} - 6(\sqrt{3})^3 + 45\sqrt{3} \right) \right]$$<BR>\n" );
document.write( "$$V_{x=9} = \pi \left[ \left( \frac{243}{5} - 162 + 135 \right) - \left( \frac{9\sqrt{3}}{5} - 18\sqrt{3} + 45\sqrt{3} \right) \right]$$<BR>\n" );
document.write( "$$V_{x=9} = \pi \left[ \left( \frac{243}{5} - 27 \right) - \left( 27\sqrt{3} + \frac{9\sqrt{3}}{5} \right) \right]$$<BR>\n" );
document.write( "$$V_{x=9} = \pi \left[ \left( \frac{243 - 135}{5} \right) - \left( \frac{135\sqrt{3} + 9\sqrt{3}}{5} \right) \right]$$<BR>\n" );
document.write( "$$V_{x=9} = \pi \left[ \frac{108}{5} - \frac{144\sqrt{3}}{5} \right]$$<BR>\n" );
document.write( "$$V_{x=9} = \mathbf{\frac{108\pi}{5} - \frac{144\pi\sqrt{3}}{5} \text{ units}^3}$$ </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );