document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1175395>Question 1175395</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A spheroid (or oblate spheroid) is the surface obtained by rotating an<BR>\n" );
document.write( "ellipse around its minor axis. The bowl in Figure 1.41 is in the shape of<BR>\n" );
document.write( "the lower half of a spheroid; that is, its horizontal cross sections are<BR>\n" );
document.write( "circles while its vertical cross sections that pass through the center are<BR>\n" );
document.write( "semi-ellipses. If this bowl is 10 in wide at the opening and V10 in deep<BR>\n" );
document.write( "at the center, how deep does a circular cover with diameter 9 in go into<BR>\n" );
document.write( "the bowl?​ </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850618 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.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=850618\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's break down this problem step-by-step.\r<BR>\n" );
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document.write( "**1. Define the Ellipse**\r<BR>\n" );
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document.write( "* The bowl is a lower half of a spheroid, so its vertical cross-sections are semi-ellipses.<BR>\n" );
document.write( "* The bowl is 10 inches wide at the opening, so the major axis of the ellipse is 10 inches. Thus, the semi-major axis (a) is 5 inches.<BR>\n" );
document.write( "* The bowl is 10 inches deep at the center, so the semi-minor axis (b) is 10 inches.<BR>\n" );
document.write( "* The equation of the ellipse, centered at the origin, is:<BR>\n" );
document.write( "    * x²/a² + y²/b² = 1<BR>\n" );
document.write( "    * x²/5² + y²/10² = 1<BR>\n" );
document.write( "    * x²/25 + y²/100 = 1\r<BR>\n" );
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document.write( "**2. Relate the Circular Cover to the Ellipse**\r<BR>\n" );
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document.write( "* The circular cover has a diameter of 9 inches, so its radius (r) is 4.5 inches.<BR>\n" );
document.write( "* We want to find the depth (y) at which the cover fits into the bowl.<BR>\n" );
document.write( "* When the cover fits into the bowl, the x-coordinate of the ellipse is equal to the radius of the circular cover.<BR>\n" );
document.write( "* Therefore, x = 4.5 inches.\r<BR>\n" );
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document.write( "**3. Solve for the Depth (y)**\r<BR>\n" );
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document.write( "* Substitute x = 4.5 into the ellipse equation:<BR>\n" );
document.write( "    * (4.5)²/25 + y²/100 = 1<BR>\n" );
document.write( "    * 20.25/25 + y²/100 = 1<BR>\n" );
document.write( "    * 0.81 + y²/100 = 1<BR>\n" );
document.write( "    * y²/100 = 1 - 0.81<BR>\n" );
document.write( "    * y²/100 = 0.19<BR>\n" );
document.write( "    * y² = 19<BR>\n" );
document.write( "    * y = √19<BR>\n" );
document.write( "    * y ≈ 4.3589 inches\r<BR>\n" );
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document.write( "**4. Calculate the Depth from the Top**\r<BR>\n" );
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document.write( "* The total depth of the bowl is 10 inches.<BR>\n" );
document.write( "* The y-coordinate we found (√19) is the distance from the bottom of the bowl.<BR>\n" );
document.write( "* The depth from the top of the bowl is:<BR>\n" );
document.write( "    * 10 - √19 ≈ 10 - 4.3589 ≈ 5.6411 inches\r<BR>\n" );
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document.write( "**Answer:**\r<BR>\n" );
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document.write( "The circular cover with a diameter of 9 inches goes approximately 5.64 inches deep into the bowl.<BR>\n" );
document.write( " </SPAN>\n" );
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