document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=752371>Question 752371</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Ok so here's the question: i have to consider the region R = { (x,y) : x is greater than or equal to 0 and less than or equal to 6 AND y is less than or equal to (2/3)x and greater than or equal to 0 }. Which i know is a triangle. Then i have to sketch the solid of revolution S obtained by revolving R about the x-axis. Which i did. Now here's where i'm stuck. I have to find the formula for the area of the cross-section perpendicular to the x-axis at x=a, which i know will be a circle. So i started out with the formula for area of a circle which is A= pi(r)^2. and i think that i somehow have to use (2/3)x and x=a to solve for r, and then substitute for the \"r\" in the equation for area so that all i have to do when given an x value later is plug it in for a, but i'm lost as to how to do that...any idea? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #457730 by </B><A HREF=/tutors/aboutme.mpl?userid=KMST><B>KMST(5396)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=KMST><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=KMST><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=457730\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Your idea is good, but to convince yourself, you need a drawing or two (or a very good imagination). Here are my drawings.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28300%2C300%2C-2%2C8%2C-2%2C8%2C%0D%0Agrid%280%29%2Cgreen%28line%28-3%2C-2%2C9%2C6%29%29%2C%0D%0Alocate%286%2C6.5%2Cgreen%28y=%282%2F3%29x%29%29%2C%0D%0Atriangle%280%2C0%2C6%2C4%2C6%2C0%29%2C%0D%0Alocate%283.5%2C1.5%2CR%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Let's loox at the x-y plane, and consider points in region R that at part of that cross section at <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=a ALIGN=MIDDLE  ALT=\"x=a\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "They form segment AP, that goes from the x-axis to that green line.<BR>\n" );
document.write( "Points A and P have <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=a ALIGN=MIDDLE  ALT=\"x=a\"   DISABLED_style= \"max-width:90%; height:auto;\" >. The radius of the circle is AP <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28300%2C300%2C-2%2C8%2C-2%2C8%2C%0D%0Arectangle%285.8%2C0%2C6%2C0.2%29%2C%0D%0Ared%28line%28-2%2C0%2C8%2C0%29%29%2Cred%28line%280%2C-2%2C0%2C8%29%29%2C%0D%0Ared%28arrow%285%2C0%2C8%2C0%29%29%2Cred%28arrow%280%2C5%2C0%2C8%29%29%2C%0D%0Alocate%287.6%2C-0.2%2Cx%29%2Clocate%28-0.5%2C7.9%2Cy%29%2C%0D%0Agreen%28line%28-3%2C-2%2C9%2C6%29%29%2C%0D%0Alocate%286%2C6.5%2Cgreen%28y=%282%2F3%29x%29%29%2C%0D%0Ablue%28line%286%2C3.8%2C6%2C0.2%29%29%2C%0D%0Alocate%287.5%2C2.5%2CR%29%2C%0D%0Aline%286%2C-0.2%2C6%2C0.2%29%2Clocate%285.9%2C-0.3%2Ca%29%2Clocate%286.1%2C0.5%2CA%29%2C%0D%0Acircle%286%2C4%2C0.2%29%2Clocate%286.1%2C4.6%2CP%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Since P is on the line, its y-coordinate is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%282%2F3%29a ALIGN=MIDDLE  ALT=\"%282%2F3%29a\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and that is the distance AP, the radius of the circle.<BR>\n" );
document.write( "So <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=pi%2A%282a%2F3%29%5E2 ALIGN=MIDDLE  ALT=\"A=pi%2A%282a%2F3%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > --> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=%284a%5E2%2F9%29%2Api ALIGN=MIDDLE  ALT=\"A=%284a%5E2%2F9%29%2Api\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=4a%5E2%2Api%2F9 ALIGN=MIDDLE  ALT=\"A=4a%5E2%2Api%2F9\"   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
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