##DESCRIPTION
## Volume of revolution of parabolic region around x-axis
## Student must setup the integral, entering limits and integrand,
## then give the numerical value.
## Limits weighted 20%, integrand and answer weighted 50/30.
## These percentages can be adjusted if desired.
##ENDDESCRIPTION

## DBsubject(Calculus - single variable)
## DBchapter(Applications of integration)
## DBsection(Volumes by disks)
## Institution(Agnes Scott College)
## Author(Larry Riddle)
## Answer boxes for limits of integration Coding in PG - Nathan Wallach (CSS based formatting work and more)
## Answer boxes for limits of integration Coding in PGML - Glenn Rice
## https://webwork.maa.org/moodle/mod/forum/discuss.php?d=4767#p14157
## Level(2)
## MO(1)
## TitleText1('APEX Calculus')
## AuthorText1('Hartman')
## EditionText1('4.0')
## Section1('7.2')
## Problem1('4')

DOCUMENT();

loadMacros(
  "PGstandard.pl",
  "MathObjects.pl",
  "PGML.pl",
  "PGgraphmacros.pl",
  "weightedGrader.pl",
  "contextFraction.pl",
  "answerHints.pl",
  "PGcourse.pl"
);

install_weighted_grader();

TEXT(beginproblem());

$showPartialCorrectAnswers = 1;
$refreshCachedImages=1;

Context("Numeric")->flags->set(
formatStudentAnswer => parsed,
reduceConstantFunctions => 0,);

$a = random(4,16);
$xlow = ($a==4 or $a==9 or $a==16) ? Real(-sqrt($a)) : Formula("-sqrt($a)");
$xhigh = ($a==4 or $a==9 or $a==16) ? Real(sqrt($a)) : Formula("sqrt($a)");
$s = sqrt($a);
$f = Formula("$a-x^2");
$integrand = Formula("pi*($a-x^2)^2");
$ans = Compute("16/15*pi*$a^(5/2)");
$rans = Real("16/15*pi*$a^(5/2)");

$xmin = floor(-sqrt($a))-1;
$xmax = ceil(sqrt($a))+1;
$ymin = -1;
$ymax = $a+1;

 
$gr = init_graph($xmin,$ymin,$xmax,$ymax,axes=>[0,0], ticks=>[$xmax-$xmin,$ymax-$ymin], size=>[400,400]);
$gr->lb('reset');
$gr->new_color("lightblue", 214,230,244); # RGB
$gr->new_color("darkblue",  100,100,255);
add_functions($gr, "$f for x in [-$s,$s] using color:darkblue and weight:2");
$gr->moveTo($xmin,0);
$gr->lineTo($xmax,0,"darkblue",1);
$gr->fillRegion([0.5,0.5,"lightblue"]);

$i = 0;           # Number the axes
do {
  $xtick = $i;
  $labelx1 = new Label($xtick,-0.2, "$xtick",'black','center');
  $labelx2 = new Label(-$xtick,-0.2, "-$xtick",'black','center');
  if ($xtick!=0) {
     $gr->lb($labelx1);
     $gr->lb($labelx2);
  } 
  $i =$i+1;
} while ($i<($xmax-$xmin)-1);

$i = 0;
do {
  $ytick = $i;
  $labely = new Label(-0.2,$ytick, "$ytick",'black', 'right','middle');
  if ($ytick!=0) {$gr->lb($labely);} 
  $i =$i+1;
} while ($i<($ymax-$ymin)-1);

# Code to format the answer boxes for integration limits
TEXT( MODES(
  HTML=>"
    <style>
      .lowerIntegrationBoundOfPair input[type=text].codeshard {
        padding:1px;
        font-size:11pt;
        height:20px !important;
      }
      .upperIntegrationBoundOfPair input[type=text].codeshard {
        padding:1px;
        font-size:11pt;
        height:20px !important;
      }
      .divOnLineWithIntegrationLimits {
        display:inline-block;
        padding-top: 15px;
        position: relative;
        left: 0px;
      }
      .divIntegrand {
        display:inline-block;
        position: relative;
        left: -8px;
      }
      .gridForPairOfIntegrationBounds {
        display:inline-grid;
        position: relative;
        top: -17px;
        left: -6px;
        grid-gap: 6px;
        text-align: left;
      }
      .lowerIntegrationBoundOfPair {
        grid-column: 1; grid-row: 2;
      }
      .upperIntegrationBoundOfPair {
        grid-column: 1; grid-row: 1;
        padding-left: 10px;
      }
   </style>",
    TeX=>""
));

# ===============================================================

#  Display the answer blanks properly in different modes

Context()->texStrings;
if ($displayMode eq 'TeX') {
   $integral1 = join("", (
      '\( \displaystyle \text{Volume = } \int_{ ',
      $xlow->ans_rule(5),
      ' }^{ ',
      $xhigh->ans_rule(5),
      '}\)', $xhigh->ans_rule(30), '\(\, dx \)' ));
  } else {
   $integral1 = join("", (
        openDiv( { "class" => "divOnLineWithIntegrationLimits" } ),
          '\( \displaystyle \quad\quad \text{Volume = } \int \)',
        closeDiv(),

        openDiv( {  "class" => "gridForPairOfIntegrationBounds" } ),
          openDiv( {  "class" => "lowerIntegrationBoundOfPair" } ),
            $xlow->ans_rule(8),
          closeDiv(),
          openDiv( {  "class" => "upperIntegrationBoundOfPair" } ),
            $xhigh->ans_rule(8),
          closeDiv(),
        closeDiv(),

        openDiv( { "class" => "divIntegrand" } ),
           $integrand->ans_rule(15),
          '\( \, dx \)',
        closeDiv(),
   ) );
}
Context()->normalStrings;


BEGIN_PGML

>>[@ image(insertGraph($gr),width=>300, height=>300, tex_size=>400) @]*<<
>>[`\quad`]Graph of [`y = [$a] - x^2`]<<

a) A region of the Cartesian plane is shaded, as shown in the figure above. Use the Disk/Washer Method to set up an integral that gives the volume of the solid of revolution formed by revolving the  region around the [`x`]-axis.  
[`\;`]  
[@ openDiv() . $integral1 . closeDiv() @]*

b) Compute the volume of the solid. [_______________]

END_PGML


WEIGHTED_ANS( $xlow->cmp,10  );
WEIGHTED_ANS( $xhigh->cmp ,10);
WEIGHTED_ANS( $integrand->cmp->withPostFilter(AnswerHints( 
  Formula("$f") => "This looks like you are finding the area of the region, not the volume of the solid of revolution.",
 [Formula("pi*$f"),Formula("($f)^2")] => "This is almost correct. What is the formula for the area of a circle?",  
)),50 );
WEIGHTED_ANS( $rans->cmp,30 );

Context("Fraction")->flags->set(reduceConstantFunctions => 0,);
$a2 = $a*$a;
$a3 = Fraction(2*$a,3);
$c1 = Fraction(1,5);
$c2 = Fraction(8,15);
$c3 = Fraction(16,15);
if ($a==4 or $a==9 or $a==16) {
  $w1 = $c2*sqrt($a)**5;
  $w2 = $c3*sqrt($a)**5;
  $c2 = "";
  $c3 = "";
} else {
  $w1 = "($a)^{5/2}";
  $w2 = "($a)^{5/2}";
}
  


BEGIN_PGML_SOLUTION

The curve [`y = [$f]`] intersects the [`x`]-axis where [`[$f]=0`], so at [`x = [$xlow]`] and [`x = [$xhigh]`]. A slice of the region rotated around the [`x`]-axis forms a disk of radius [`R(x) = [$f]`]. The cross-sectional area of the slice is [`\pi R(x)^2 = \pi([$f])^2`]. The volume is therefore given by the integral

[``\int_{[$xlow]}^{[$xhigh]}\pi R(x)^2\;dx =  \int_{[$xlow]}^{[$xhigh]}\pi([$f])^2\;dx \approx [$ans]``]

Here is the calculation with the antidifferentiation carried out.

[``\begin{aligned}
\int_{[$xlow]}^{[$xhigh]}\pi([$f])^2\;dx &= \pi \int_{[$xlow]}^{[$xhigh]} \left([$a2]-[$a*2]x^2 + x^4\right)\;dx \\ \\
&= \pi \left([$a2]x - [$a3]x^3 + [$c1]x^5\right)\bigg|_{[$xlow]}^{[$xhigh]} \\ \\
&= \pi \left([$c2][$w1] - (-[$c2][$w1]\right) = [$c3][$w2]\pi
\end{aligned}
``]

END_PGML_SOLUTION

COMMENT('Randomization provides 13 different possible versions of this question.');

ENDDOCUMENT();