document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1155797>Question 1155797</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let R be the circle centered at (0,0) with radius 10. The lines x = 6 and y = 5 divide R into four regions R1, R2, R3, and R4. Let R denote the area of region Ri. If R1 > R2 > R3 > R4,then find R1 - R2 - R3 + R4. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #778533 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=778533\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Clearly the largest region R1 is below y=5 and left of x=6.<br><BR>\n" );
document.write( "Clearly the smallest region R4 is above y=5 and right of x=6.<br><BR>\n" );
document.write( "The other two regions are (a) above y=5 and left of x=6 and (b) below y=5 and right of x=6.  Since x=6 is farther from the center of the circle than y=5, R2 is the region above y=5 and left of x=6 and R3 is the region below y=5 and right of x=6.<br><BR>\n" );
document.write( "R2 and R4 together are the portion of the circle above y=5; R3 and R4 together are the portion right of x=6.  There are at least two ways to find the areas of those combined regions.<br><BR>\n" );
document.write( "Let's look at those two ways for finding the area of R2+R4.<br><BR>\n" );
document.write( "(1) calculating the area with calculus....<br><BR>\n" );
document.write( "The y-axis divides this region into two congruent parts; so let's find the area of one of those halves.<br><BR>\n" );
document.write( "The intersection of the line y=5 and the right half of the circle is (5sqrt(3),5).<br><BR>\n" );
document.write( "The area of this region is then<br><BR>\n" );
document.write( "(integral from 0 to 5sqrt(3) of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28100-x%5E2%29 ALIGN=MIDDLE  ALT=\"sqrt%28100-x%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >), minus the area of the rectangle bounded by x=0, y=0, x=5sqrt(3), and y=5.<br><BR>\n" );
document.write( "Or, equivalently, that area is<br><BR>\n" );
document.write( "integral from 0 to 5sqrt(3) of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28100-x%5E2%29-5 ALIGN=MIDDLE  ALT=\"sqrt%28100-x%5E2%29-5\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "A table of integrals show the integral of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28100-x%5E2%29 ALIGN=MIDDLE  ALT=\"sqrt%28100-x%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > is<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%29%28x%29%28sqrt%28100-x%5E2%29%29%2B50sin%5E%28-1%29%28x%2Fsqrt%28100-x%5E2%29%29 ALIGN=MIDDLE  ALT=\"%281%2F2%29%28x%29%28sqrt%28100-x%5E2%29%29%2B50sin%5E%28-1%29%28x%2Fsqrt%28100-x%5E2%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Plugging in x=5sqrt(3) and using that integral, the area of R2 and R4 together is<br><BR>\n" );
document.write( "R2+R4 = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%28%281%2F2%29%285sqrt%283%29%29%285%29%2B50sin%5E%28-1%29%28sqrt%283%29%2F2%29-5%285sqrt%283%29%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "(2) calculating the area as a sector of the circle minus a triangle....<br><BR>\n" );
document.write( "The area of R2 and R4 together is the area of the sector of the circle defined by the line y=5, minus the area of the triangle defined by the center of the circle and the line y=5.  Using the angles of the triangle to determine what fraction of a circle we have, that area is<br><BR>\n" );
document.write( "R2+R4 = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%282sin%5E%28-1%29%28sqrt%283%29%2F2%29%2F2pi%29%28100pi%29-5%285sqrt%283%29%29 ALIGN=MIDDLE  ALT=\"%282sin%5E%28-1%29%28sqrt%283%29%2F2%29%2F2pi%29%28100pi%29-5%285sqrt%283%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "That is exactly the same calculation we had using calculus....<br><BR>\n" );
document.write( "You can find the combined areas of R3 and R4 in the same manner.  The calculations are<br><BR>\n" );
document.write( "R3+R4 = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%28%281%2F2%29%286%29%288%29%2B50sin%5E%28-1%29%284%2F5%29-6%288%29%29 ALIGN=MIDDLE  ALT=\"2%28%281%2F2%29%286%29%288%29%2B50sin%5E%28-1%29%284%2F5%29-6%288%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "or<br><BR>\n" );
document.write( "R3+R4 = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%282sin%5E%28-1%29%284%2F5%29%2F2pi%29%28100pi%29-6%288%29 ALIGN=MIDDLE  ALT=\"%282sin%5E%28-1%29%284%2F5%29%2F2pi%29%28100pi%29-6%288%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "So now we have the areas of R2 and R4 together, and of R3 and R4 together.  To find the areas separately, we have to use calculus to find the area of R4 alone.  That area is<br><BR>\n" );
document.write( "integral from 6 to 5sqrt(3) of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28100-x%5E2%29-5 ALIGN=MIDDLE  ALT=\"sqrt%28100-x%5E2%29-5\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%28%281%2F2%29%285sqrt%283%29%29%285%29%2B50sin%5E%28-1%29%28sqrt%283%29%2F2%29%29-%282%281%2F2%29%286%29%288%29%2B50sin%5E%28-1%29%284%2F5%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "You now have all you need to find what the question is asking for -- the value of R1-R2-R3+R4:<BR>\n" );
document.write( "subtract R4 from (R2+R4) to get R2<BR>\n" );
document.write( "subtract R4 from (R3+R4) to get R3<BR>\n" );
document.write( "subtract (R2+R3+R4) from 100pi to get R1<BR>\n" );
document.write( "and finally calculate R1-R2-R3+R4<br><BR>\n" );
document.write( "Note since R1+R2+R3+R4=100pi, you can also calculate R1-R2-R3+R4 as 100pi-2(R2+R3).<br><BR>\n" );
document.write( "I leave all the ugly calculations to you....<br><BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );