![](\"resources/question-resources/malteser.png\")
$4\\pi^2 r \\Delta r$
", "$\\frac{4}{3}\\pi r^3 \\Delta r$
", "$4\\pi r^2 \\Delta r$
", "$2\\pi r \\Delta r$
"], "variableReplacements": [], "displayColumns": 0, "scripts": {}, "shuffleChoices": false, "maxMarks": 0, "marks": 0, "prompt": "We consider the mass of a thin spherical shell of thickness $\\Delta r$ and radius $r$. The volume of the thin shell is approximately its surface area times its thickness. Which is the correct expression for the volume of a single thin shell?
", "variableReplacementStrategy": "originalfirst", "minMarks": 0, "distractors": ["", "", "", ""]}, {"showCorrectAnswer": true, "showFeedbackIcon": true, "displayType": "radiogroup", "matrix": [0, 0, "1", 0], "type": "1_n_2", "choices": ["$M=\\sum\\limits_{i=0}^N4\\pi^2 r_i \\rho_i \\Delta r$
", "$M=\\sum\\limits_{i=0}^N\\frac{4}{3}\\pi r^3 \\rho_i\\Delta r$
", "$M=\\sum\\limits_{i=0}^N4\\pi r^2 \\rho_i\\Delta r$
", "$M=\\sum\\limits_{i=0}^N2\\pi r \\rho_i\\Delta r$
"], "variableReplacements": [], "displayColumns": 0, "scripts": {}, "shuffleChoices": false, "maxMarks": 0, "marks": 0, "prompt": "To find the mass of a shell we need to multiply the volume by the density. We split the planet into $N$ thin shells and call $\\rho_i$ the density of the $i$th shell. By summing over all the shell volumes we obtain the volume of the entire planet. Which of the following summations gives the correct value of the mass, $M$?
", "variableReplacementStrategy": "originalfirst", "minMarks": 0, "distractors": ["", "", "", ""]}, {"variableReplacements": [], "scripts": {}, "marks": 0, "gaps": [{"showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "variableReplacements": [], "scripts": {}, "type": "numberentry", "correctAnswerStyle": "plain", "allowFractions": false, "minValue": "0", "maxValue": "0", "marks": 1, "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showFeedbackIcon": true}, {"showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "variableReplacements": [], "scripts": {}, "type": "numberentry", "correctAnswerStyle": "plain", "allowFractions": false, "minValue": "{radius}*1000", "maxValue": "{radius}*1000", "marks": 1, "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showFeedbackIcon": true}], "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "type": "gapfill", "prompt": "This summation estimate of the mass increases in accuracy with more and more shells of smaller and smaller thickness. If we have an infinite number of infinitely thin shells, our summation can be written as an integral, with $\\Delta r$ becoming $\\text{d}r$. If we assume a constant density of Planet Malteser, what are the limits of integration in metres?
\nr = [[0]] to [[1]]
"}, {"showCorrectAnswer": true, "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": 0, "gaps": [{"showCorrectAnswer": true, "checkingaccuracy": 0.001, "answer": "{siground(4pi/3 * ({density[0]}*(r1*1000)^3 + {density[1]}*((radius*1000)^3 -(r1*1000)^3)),3)}", "showFeedbackIcon": true, "vsetrange": [0, 1], "showpreview": true, "type": "jme", "checkingtype": "absdiff", "variableReplacements": [], "scripts": {}, "vsetrangepoints": 5, "marks": 1, "checkvariablenames": false, "variableReplacementStrategy": "originalfirst", "expectedvariablenames": []}], "prompt": "The density of the honeycomb core is $\\var{density[0]}$ kg m-3 and the density of the milk chocolate mantle is $\\var{density[1]}$ kg m-3. Given that the diameter of the planet is $\\var{radius*2}$ km and the core/mantle boundary radius is at $\\var{r1}$ km, calculate the mass of Planet Malteser to 3 significant figures. You may enter an answer in standard form e.g. 4*10^3 = $4\\times10^3$. Also remember to use consistent units in your calculations.
\nMass = [[0]] kg
", "stepsPenalty": 0, "showFeedbackIcon": true, "type": "gapfill", "variableReplacements": [], "steps": [{"variableReplacements": [], "scripts": {}, "marks": 1, "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "type": "extension", "prompt": "Because the densities are different between the core and the mantle, you need to do two definte integration. one is from the centre to the edge of the core and the other is from the edge of the core to the edge of the mantle. see the advice for more detail.
\n"}]}], "name": "10: Integration - Summation of Planet", "ungrouped_variables": ["radius", "r1", "density"], "statement": "Planet Malteser has a honeycomb core and a milk chocolate mantle. It has a diameter of $\\var{2*radius}$ km. The core/mantle boundary radius is at $\\var{r1}$ km.
\n
Figure 1 A cross-section of Planet Malteser
\nIn the following parts of this question you will estimate the total mass of Planet Malteser.
", "variables": {"r1": {"description": "", "definition": "random(1300..1800#100)", "name": "r1", "group": "Ungrouped variables", "templateType": "anything"}, "density": {"description": "", "definition": "[random(500..1000#10),random(1000..2000#10)]", "name": "density", "group": "Ungrouped variables", "templateType": "anything"}, "radius": {"description": "", "definition": "2000", "name": "radius", "group": "Ungrouped variables", "templateType": "anything"}}, "rulesets": {}, "metadata": {"description": "This question is application of integration in geology using the concept of summation.
", "licence": "All rights reserved"}, "tags": [], "extensions": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "functions": {}, "variable_groups": [], "preamble": {"css": "", "js": ""}, "advice": "a) The formula for the surface area of a sphere is $4\\pi r^2$. For a suitably small thickness $\\Delta r$ we approximate the volume of a shell by $4\\pi r^2 \\Delta r$. To find the volume of a considerably thicker shell we would have to instead use the formula for the volume of a sphere.
\nb) Mass is density times volume, so we multiply the previous formula by the density $\\rho$. By labelling $r$ and $\\rho$ with $i$, we can calculate the mass of a spherical object where its density is dependent on the distance from the centre.
\nc) Assuming a constant average density of the planet, we would integrate from $r= 0$ m to $r = \\var{radius*1000}$ m, (half of the diameter, {radius*2} km, converted into metres).
\nd) We need to calculate the integrals in the expression $\\int\\limits_0^b \\left( 4\\pi r^2\\rho_c \\right) \\text{d} r +\\int\\limits_b^{d/2}\\left( 4\\pi r^2\\rho_m \\right) \\text{d} r$, where $\\rho_c$ is the density of the core, $\\rho_m$ is the density of the mantle, $b$ is the core-mantle boundary radius and $d$ is the diameter of the planet. Since our densities are given in S.I. units we need to convert any distances we use into metres from kilometres. We can then continue with the calculations:
\n$\\begin{array}{rcl}M &=& 4\\pi\\left(\\rho_c\\int\\limits_0^{\\var{r1*1000}} r^2 \\space\\text{d}r + \\rho_m\\int\\limits_{\\var{r1*1000}}^{\\var{radius*1000}} r^2 \\space\\text{d}r \\right)\\\\&=&4\\pi\\left(\\rho_c\\left[\\frac{r^3}{3}\\right]_0^{\\var{r1*1000}} + \\rho_m\\left[\\frac{r^3}{3}\\right]_{\\var{r1*1000}}^{\\var{radius*1000}}\\right)\\\\&=&4\\pi\\left(\\var{density[0]}\\times\\left(\\frac{\\var{r1*1000}^3}{3} - 0\\right) +\\var{density[1]}\\times\\left(\\frac{\\var{radius*1000}^3}{3} - \\frac{\\var{r1*1000}^3}{3}\\right)\\right)\\\\&=& \\var{4*pi/3 * (density[0]*(r1*1000)^3 + (density[1])*((radius*1000)^3 -(r1*1000)^3))}\\\\&=&\\var{siground(4*pi/3 * (density[0]*(r1*1000)^3 + (density[1])*((radius*1000)^3 -(r1*1000)^3)),3)}\\space\\text{kg to 3 significant figures}\\end{array}$
", "type": "question", "contributors": [{"name": "John Hodkinson", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/921/"}]}]}], "contributors": [{"name": "John Hodkinson", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/921/"}]}