// Numbas version: exam_results_page_options {"name": "Double Integrals", "metadata": {"description": "

set of exercises on double and triple integrals

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Calculate a repeated integral of the form $\\displaystyle I=\\int_0^1\\;\\int_0^{x^{m-1}}mf(x^m+a)dy \\;dx$

The $y$ integral is trivial, and the $x$ integral is of the form $g'(x)f'(g(x))$, so it straightforwardly integrates to $f(g(x))$.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following repeated integral:

\\[ I = \\int_0^1 \\; \\int_0^{\\simplify[all]{x^{m-1}}}\\var{m} \\var{fun} \\; \\mathrm{d}y\\; \\mathrm{d}x \\]

", "advice": "

We want to find

\\[ I=\\int_0^1 \\; \\int_0^{\\simplify[all]{x^{m-1}}} \\var{m} \\var{fun} \\; \\mathrm{d}y \\; dx \\]

The innermost integral gives:

\\[ \\int_0^{\\simplify[all]{x^{m-1}}}\\var{m} \\var{fun} \\; \\mathrm{d}y = \\left[\\var{m}y \\; \\var{fun} \\right]_0^{\\simplify[all]{x^{m-1}}}=\\simplify[all]{{m}x^{m-1}} \\var{fun} \\]

So we have to find $\\displaystyle I=\\int_0^1\\simplify[all]{{m}x^{m-1}} \\var{fun} \\; \\mathrm{d}x$.

Note that if we use the substitution $u=\\simplify[all]{x^{m}+{a}}$ then it is easy to find this last definite integral and we find that:

$I=\\var{ans}$ to 3 decimal places.

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$I=\\;$[[0]]

Input your answer to 3 decimal places.

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3 Repeated integrals of the form $\\int_a^b\\;\\int_c^{f(x)}g(x,y)\\;dy \\;dx$ where $g(x,y)$ is a polynomial in $x,\\;y$ and $f(x)$ is a degree 0, 1 or 2 polynomial in $x$.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Calculate the following repeated integrals.

", "advice": "

a)

\\[I = \\int_0^{\\var{a}} \\;\\int_{\\var{f}}^{\\var{g}}\\simplify[all]{({b}+{c}*x+{d}*y)} \\; \\mathrm{d}y\\; \\mathrm{d}x\\]

Calculating the inner integral, we have:

\\begin{align}
\\int_{\\var{f}}^{\\var{g}}\\simplify[all,!noleadingminus,!collectNumbers]{({b}+{c}*x+{d}*y)} \\; \\mathrm{d}y &= \\left[\\simplify[all,!noleadingminus,!collectNumbers]{{b}y+{c}*x*y+{d}*y^2/2}\\right]_{\\var{f}}^{\\var{g}}\\\\
&= \\simplify[all,!noleadingminus,!collectNumbers]{{b} * {g} + {c} * {g} * x + {d / 2} * {g ^ 2} + {b} * { -f} + {c} * { -f} * x + {d / 2} * { -(f ^ 2)}} \\\\
&= \\simplify[all,!noleadingminus,!collectNumbers]{ {b * g -(b * f) + (d / 2) * (g ^ 2 -(f ^ 2))} + {c * g -(c * f)} * x}
\\end{align}

The outer integral gives:

\\begin{align}
I &= \\simplify[std]{DefInt({b * g -(b * f) + (d / 2) * (g ^ 2 -(f ^ 2))} + {c * g -(c * f)} * x,x,0,{a}) } \\\\
&= \\left[\\simplify[std]{{b * g -(b * f) + (d / 2) * (g ^ 2 -(f ^ 2))} * x + {(c * g -(c * f)) / 2} * x ^ 2}\\right]_0^{\\var{a}} \\\\
&= \\var{ans1}
\\end{align}

b)

\\[ I=\\var{(m + 1) * (m + n + 2)} \\int_0^1 \\; \\mathrm{d}x \\int_x^{\\var{b1}}\\simplify[std]{({n + 1} * x ^ {m} * y ^ {n})} \\; \\mathrm{d}y \\]

Calculating the inner integral, we have:

\\begin{align}
\\int_x^{\\var{b1}}\\simplify[std]{({n + 1} * x ^ {m} * y ^ {n})}dy &= \\left[x^{\\var{m}}y^{\\var{n+1}}\\right]_x^{\\var{b1}} \\\\
&= \\simplify{{b1 ^ (n + 1)}* x ^ {m} -(x ^ {m + n + 1})}
\\end{align}

Finally the outer integral gives:

\\begin{align}
I &= \\var{(m + 1) * (m + n + 2)}\\int_0^1\\simplify[std]{{b1^ (n + 1)} * x ^ {m} -(x ^ {m + n + 1})} \\; \\mathrm{d}x \\\\
&= \\simplify[std]{ {(m + 1) * (m + n + 2)} * ({b1 ^ (n + 1)} / {m + 1} -(1 / {m + n + 2})) } \\\\
&= \\var{ans2}
\\end{align}

c)

\\[ I=\\var{con}\\int_{-1}^1 \\; \\int_0^{\\simplify[all]{{c2}+{d2}*x^{p2}}}\\simplify[all]{{c1}+{d1}*y^{p1}} \\; \\mathrm{d}y \\; \\mathrm{d}x \\]

Calculating the inner integral, we have:

\\begin{align}
\\int_0^{\\simplify[all]{{c2}+{d2}*x^{p2}}}\\simplify[all]{{c1}+{d1}*y^{p1}} \\; \\mathrm{d}y &= \\left[\\simplify[all]{{c1} * y + {d1 / (p1 + 1)} * y ^ {p1 + 1}}\\right]_0^{\\simplify[all]{{c2}+{d2}*x^{p2}}} \\\\
&= \\simplify[std]{{c1} * ({c2} + {d2} * x ^ {p2}) + {d1 / (p1 + 1)} * ({c2} + {d2} * x ^ {p2}) ^ {p1 + 1}} \\\\
&= \\simplify[std]{ {c1 * c2} + {c1 * d2} * x ^ {p2} + {p1 -1} * {h1} * ({c2 ^ 3} + {3 * c2 ^ 2 * d2} * x ^ {p2} + {3 * c2 * d2 ^ 2} * x ^ {2 * p2} + {d2} ^ 3 * x ^ {3 * p2}) + {2 -p1} * {h1} * ({c2 ^ 2} + {2 * c2 * d2} * x ^ {p2} + {d2 ^ 2} * x ^ {2 * p2})} \\\\
&= \\simplify[std,collectNumbers]{{c1 * c2 + (p1 -1) * h1 * c2 ^ 3 + (2 -p1) * h1 * c2 ^ 2} + {c1 * d2 + (p1 -1) * h1 * 3 * c2 ^ 2 * d2 + (2 -p1) * h1 * 2 * c2 * d2} * x ^ {p2} + {(p1 -1) * h1 * 3 * c2 * d2 ^ 2 + (2 -p1) * h1 * d2 ^ 2} * x ^ {2 * p2} + {(p1 -1) * h1 * d2 ^ 3} * x ^ {3 * p2}}
\\end{align}

Finally the outer integral gives:

\\begin{align}
I &= \\simplify[std]{{con} * DefInt({c1 * c2 + (p1 -1) * h1 * c2 ^ 3 + (2 -p1) * h1 * c2 ^ 2} + {c1 * d2 + (p1 -1) * h1 * 3 * c2 ^ 2 * d2 + (2 -p1) * h1 * 2 * c2 * d2} * x ^ {p2} + {(p1 -1) * h1 * 3 * c2 * d2 ^ 2 + (2 -p1) * h1 * d2 ^ 2} * x ^ {2 * p2} + {(p1 -1) * h1 * d2 ^ 3} * x ^ {3 * p2},x, -1,1)} \\\\
&= \\var{ans3}
\\end{align}

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\\[I = \\int_0^{\\var{a}}\\;\\int_{\\var{f}}^{\\var{g}}\\simplify[all]{({b}+{c}*x+{d}*y)}\\;dy \\;dx\\]

$I=$ [[0]]

\\[I=\\var{(m+1)(m+n+2)}\\int _0^1\\;\\int_x^{\\var{b1}}\\simplify[all]{{n+1}*x^{m}*y^{n}}\\;dy \\;dx\\]

$I=\\;$[[0]]

\\[I=\\var{con}\\int_{-1}^1\\;\\int_0^{\\simplify[all]{{c2}+{d2}*x^{p2}}}\\simplify[all]{{c1}+{d1}*y^{p1}}\\;dy \\;dx\\]

$I=\\;$?[[0]]

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Inputing Integrals tutorial

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To input integrals in Numbas we use the commands int() and defint(). We will see how to use them in more detail here.

This is a tutorial

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To input the integral $\\int x^3 \\,dx$, write int(x^3, x) in the area below.

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One can also write the expression $\\int\\, dx$. To do so, input int('',x). The expression '' tells that the firtst argument of the function is an empty string.

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You can also input definite integrals. For this we use defint(). This function has four arguments, as made clear in the below example:

To input $\\int^b_a f(x) \\, dx$, write defint(f(x), x, a, b).

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For inputting double integrals, one can use int() and defint() nested.

For example to write $\\int^b_a\\int^d_c g(x,y) \\,dx\\, dy$, input defint(defint(g(x,y), x, c,d),y,a,b)

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Calculating the area enclosed between a linear function and a quadratic function by integration. The limits (points of intersection) are not given in the question and must be calculated.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the area enclosed by $\\simplify{y={m}x+{c_1}}$ and $\\simplify{y=x^2+{b}x+{c_2}}$ using a double integral.

{geogebra_applet('https://www.geogebra.org/m/fyxr2fqj',defs)}

", "advice": "

To find the points of intersection, we will solve the given eqautions simultaneously. We equate the two and solve for $x$ first:

\\[ \\begin{split} \\simplify{x^2+{b}x+{c_2}} &\\,= \\simplify{{m}x+{c_1}}, \\\\ \\simplify{x^2+{b-m}x+{c_2-c_1}} &\\,=0, \\\\ \\simplify{(x-{cp1})(x-{cp2})}&\\,=0. \\end{split} \\]

Therefore, the points of intersection are when $\\simplify{x1={cp1}}$ and $\\simplify{x2={cp2}}$.

Now by pluging-in these values into $\\simplify{y={m}x+{c_1}}$ we get $\\simplify{y_1 = {y1}}$ and $\\simplify{y_2 = {y2}}$.

We are asked to integrate $y$ first. So, $x$ can vary freely in the range we found above, and $y$ will vary depending on $x$. So the relevant boundaries of the variables are

\\[ \\var{x1} < x < \\var{x2} \\quad \\mbox{ and }\\quad \\simplify{x^2+{b}x+{c_2}} < y < \\simplify{{m}x+{c_1}}\\simplify{x^2+{b}x+{c_2}}.\\]

Then the integral

\\[\\int_{\\var{x1}}^{\\var{x2}} \\int_{\\simplify{x^2+{b}x+{c_2}}}^{\\simplify{{m}x+{c_1}}}\\, \\mathrm{d}y\\, \\mathrm{d}x\\]

can be used to compute the shaded area.

We start with the inner integral first

\\[\\int_{\\simplify{x^2+{b}x+{c_2}}}^{\\simplify{{m}x+{c_1}}}\\, \\mathrm{d}y = y|_{\\simplify{x^2+{b}x+{c_2}}}^{\\simplify{{m}x+{c_1}}} = \\simplify{{m}x+{c_1}} - (\\simplify{x^2+{b}x+{c_2}}) = \\simplify{-x^2 + {m-b}x + {c_1-c_2}}.\\]

Then we get

\\[\\int_{\\var{x1}}^{\\var{x2}} \\int_{\\simplify{x^2+{b}x+{c_2}}}^{\\simplify{{m}x+{c_1}}}\\, \\mathrm{d}y\\, \\mathrm{d}x =\\] {advice}

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\\\\[ \\\\begin{split}\\\\simplify[all, !noLeadingMinus]{defint(-x^2+{m-b}x+{c_1-c_2},x,{cp1},{cp2})} &\\\\,= \\\\left[\\\\simplify[all, !noLeadingMinus]{-x^3/3+{m-b}x^2/2+{c_1-c_2}x} \\\\right]_\\\\var{cp1}^\\\\var{cp2},\\\\\\\\ &\\\\,=\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp2^3/3}+{(m-b)*cp2^2/2}+{(c_1-c_2)*cp2}}\\\\right]-\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp1^3/3}+{(m-b)*cp1^2/2}+{(c_1-c_2)*cp1}}\\\\right]\\\\\\\\ &\\\\,= \\\\left(\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2}}\\\\right)-\\\\left(\\\\simplify[all, fractionNumbers]{{-(cp1)^3/3+(m-b)cp1^2/2+(c_1-c_2)*cp1}}\\\\right)\\\\\\\\ &\\\\,=\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2+(cp1)^3/3-(m-b)cp1^2/2-(c_1-c_2)*cp1}}. \\\\end{split} \\\\]

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\\\\[ \\\\begin{split}\\\\simplify[all,!noLeadingMinus]{defint(-x^2+{m-b}x+{c_1-c_2},x,{cp1},{cp2})} &\\\\,= \\\\left[\\\\simplify[all,!noLeadingMinus]{-x^3/3+{m-b}x^2/2+{c_1-c_2}x} \\\\right]_\\\\var{cp1}^\\\\var{cp2},\\\\\\\\ &\\\\,=\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp2^3/3}+{(m-b)*cp2^2/2}+{(c_1-c_2)*cp2}}\\\\right]-\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp1^3/3}+{(m-b)*cp1^2/2}+{(c_1-c_2)*cp1}}\\\\right]\\\\\\\\ &\\\\,= \\\\left(\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2}}\\\\right)-\\\\left(\\\\simplify[all, fractionNumbers]{{-(cp1)^3/3+(m-b)cp1^2/2+(c_1-c_2)*cp1}}\\\\right)\\\\\\\\ &\\\\,=\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2+(cp1)^3/3-(m-b)cp1^2/2-(c_1-c_2)*cp1}}\\\\\\\\ &\\\\,=\\\\var{sol2dp} \\\\end{split} \\\\]

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Denote the intersection points of $\\simplify{y={m}x+{c_1}}$ and $\\simplify{y=x^2+{b}x+{c_2}}$ as $(x_1, y_1)$ (leftmost) and $(x_2,y_2)$ (rightmost). Find these points

$x_1=$[[0]], $y_1=$[[1]]

$x_2=$[[2]], $y_2=$ [[3]]

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Write a double integral, integrating $y$ first, to compute the shaded area.

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Compute the double integral you wrote in part b)

[[0]] (Give your answers to 2 decimal places where necessary)

Volume of a tetrahedron using integrals

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Find the volume of the tetrahedron bounded by the coordinate planes $x=0$, $y=0$ and $z=0$ and the plane $\\frac{x}{\\var{a}} + \\frac{y}{\\var{b}} + \\frac{z}{\\var{c}} =1$.

", "advice": "

The coordinate surfaces $x=0$, $y=0$ and $z=0$ are given. The plane $\\frac{x}{\\var{a}} + \\frac{y}{\\var{b}} + \\frac{z}{\\var{c}} = 1$ can be rewritten as a functions as follows

\\[z = f(x,y) = \\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right).\\]

The surface of $f(x,y)$ intersects the axes at $x=\\var{a}, y=\\var{b}$ and $z=\\var{c}$.

Let us integrate $y$-firts. Then the triangular region the tetrahedron sits is bounded by the lines $x=0, y=0$ and $\\frac{x}{\\var{a}} + \\frac{y}{\\var{b}} = 1$. Then we have the following bounds for integration

\\[0<x<\\var{a} \\quad \\mbox{and} \\quad 0 < y< \\frac{\\var{b}}{\\var{a}}(\\var{a} -x).\\]

Hence, we can compute the volume of the tetrahedron as

\\[V = \\int_0^\\var{a}\\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)}f(x,y) \\,dy\\, dx = \\int_0^\\var{a}\\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)} \\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right) \\,dy\\, dx =\\\\[3mm]
= \\int_0^\\var{a} \\var{c}\\left[ y - \\frac{xy}{\\var{a}} - \\frac{y^2}{\\simplify{2*{b}}}\\right]_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)}\\, dx = \\int_0^\\var{a} \\frac{\\simplify{{b*c}}}{\\simplify{{2*a^2}}}(\\var{a}-x)^2 \\, dx = \\frac{\\simplify{{a*b*c}}}{6}\\]

(enter your answer as an interger or a fraction)

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mass of tetrahedron via integration

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Cosider the tetrahedron bounded by the coordinate planes $x=0$, $y=0$ and $z=0$ and the plane $\\frac{x}{\\var{a}} + \\frac{y}{\\var{b}} + \\frac{z}{\\var{c}} =1$. Assume that it has a density function $\\rho(x) = \\var{R0}(\\var{a}+x)$.

", "advice": "

The coordinate surfaces $x=0$, $y=0$ and $z=0$ are given. The plane $\\frac{x}{\\var{a}} + \\frac{y}{\\var{b}} + \\frac{z}{\\var{c}} = 1$ can be rewritten as a functions as follows

\\[z = f(x,y) = \\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right).\\]

The surface of $f(x,y)$ intersects the axes at $x=\\var{a}, y=\\var{b}$ and $z=\\var{c}$.

We can write the boundaries of the tetrahedron as

\\[0<x<\\var{a} \\quad \\mbox{and} \\quad 0 < y< \\frac{\\var{b}}{\\var{a}}(\\var{a} -x) \\quad\\mbox{and }\\quad 0\\leq z \\leq \\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right).\\]

Thus, with mass element $dM = \\rho(x)dV = \\rho(x)dx\\,dy\\,dz$, the total mass is

\\[M= \\int_T\\,dM = \\int_0^\\var{a}\\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)}\\int_0^{\\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right)}\\rho(x)\\,dz\\,dy\\,dx =\\\\[3mm]
\\int_0^\\var{a}\\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)}\\int_0^{\\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right)} \\var{R0}(\\var{a}+x) \\,dz\\,dy\\,dx =\\\\[3mm]
\\int_0^\\var{a} \\var{R0}(\\var{a}+x) \\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)}\\int_0^{\\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right)} \\,dz\\,dy\\,dx =\\\\[3mm]
\\int_0^\\var{a} \\var{R0}(\\var{a}+x) \\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)} \\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right) \\,dy\\,dx =\\\\[3mm]
\\int_0^\\var{a} \\var{R0}(\\var{a}+x)\\left[ \\frac{\\simplify{{b*c}}}{\\simplify{{2*a^2}}}(\\var{a}-x)^2\\right]\\, dx = \\frac{5\\cdot\\var{a}^2\\cdot\\var{b}\\cdot\\var{c}\\cdot\\var{R0}}{24}=\\frac{\\simplify{{5*{a^2}*{b}*{c}*{R0}}}}{24}.\\]

We start with the $x$ coordinate.

\\[\\bar x= \\frac{1}{M}\\int_T x\\,dM = \\int_0^\\var{a}\\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)}\\int_0^{\\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right)}x\\rho(x)\\,dz\\,dy\\,dx =\\\\[3mm]
\\frac{1}{M} \\int_0^\\var{a}\\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)}\\int_0^{\\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right)} x\\var{R0}(\\var{a}+x) \\,dz\\,dy\\,dx =\\\\[3mm]
\\frac{1}{M}\\int_0^\\var{a} x\\var{R0}(\\var{a}+x) \\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)}\\int_0^{\\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right)} \\,dz\\,dy\\,dx =\\\\[3mm]
\\frac{1}{M}\\int_0^\\var{a} x\\var{R0}(\\var{a}+x) \\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)} \\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right) \\,dy\\,dx =\\\\[3mm]
\\frac{1}{M}\\int_0^\\var{a} x\\var{R0}(\\var{a}+x)\\left[ \\frac{\\simplify{{b*c}}}{\\simplify{{2*a^2}}}(\\var{a}-x)^2\\right]\\, dx =\\\\[3mm]
\\frac{\\simplify{{R0*b*c}}}{\\simplify{{2*a^2}}M}\\int_0^\\var{a} x (\\var{a}+x)\\left[ (\\var{a}-x)^2\\right]\\, dx =
\\frac{\\simplify{{R0*b*c}}}{\\simplify{{2*a^2}}M}\\int_0^\\var{a}x(\\var{a^3} -\\var{a^2}x-\\var{a} x^2 + x^3) \\, dx =\\\\[3mm]
\\frac{\\simplify{{R0*b*c}}}{\\simplify{{2*a^2}}M}\\int_0^\\var{a}\\var{a^3}x -\\var{a^2}x^2-\\var{a} x^3 + x^4 \\, dx = \\\\[3mm]
\\frac{\\simplify{{R0*b*c}}}{\\simplify{{2*a^2}}M}\\left[\\frac{\\var{a^3}x^2}{2} -\\frac{\\var{a^2}x^3}{3}-\\frac{\\var{a} x^4}{4} + \\frac{x^5}{5} \\right]_0^{\\var{a}}=\\\\[3mm]
\\simplify[fractionNumbers]{({a}^5/2 - ({a}^5)/3 - {a}^5/4 + {a}^5/5) *{R0*b*c}/{2*a^2*Mass}} =
\\simplify[fractionNumbers,otherNumbers]{({a}^5/2 - ({a}^5)/3 - {a}^5/4 + {a}^5/5) *{R0*b*c}/{2*a^2*Mass}}.\\]

Simlarly, for $y$ coordinate:

\\[\\bar y = \\frac{1}{M}\\int_T y\\,dM = \\int_0^\\var{a}\\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)}\\int_0^{\\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right)}y\\rho(x)\\,dz\\,dy\\,dx =\\\\[3mm]
\\frac{1}{M} \\int_0^\\var{a}\\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)}\\int_0^{\\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right)} y\\var{R0}(\\var{a}+x) \\,dz\\,dy\\,dx =\\\\[3mm]
\\frac{1}{M}\\int_0^\\var{a} \\var{R0}(\\var{a}+x) \\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)} y \\int_0^{\\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right)} \\,dz\\,dy\\,dx =\\\\[3mm]
\\frac{1}{M}\\int_0^\\var{a} \\var{R0}(\\var{a}+x) \\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)} y\\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right) \\,dy\\,dx =\\\\[3mm]
\\frac{1}{M}\\int_0^\\var{a} \\var{R0}(\\var{a}+x) \\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)} \\var{c}\\left( y - \\frac{xy}{\\var{a}} - \\frac{y^2}{\\var{b}}\\right) \\,dy\\,dx =\\\\[3mm]
\\frac{1}{M}\\int_0^\\var{a} \\var{R0}(\\var{a}+x)\\frac{\\simplify{{c*b^2}}}{\\simplify{{6*a^3}}}\\left[\\var{a^3}-3\\cdot\\simplify[unitFactor]{{a^2}x} +3\\cdot\\simplify[unitFactor]{{a}x^2}-x^3\\right]\\, dx =\\\\[3mm]
\\frac{\\simplify{{R0*b^2*c}}}{\\simplify{{6*a^3}}M}\\int_0^\\var{a} (\\var{a}+x)( \\var{a}-x)^3, dx =
\\frac{\\simplify{{R0*b^2*c}}}{\\simplify{{6*a^3}}M}\\int_0^\\var{a}\\var{a^4} -2\\cdot\\simplify[unitFactor]{{a^3}x} +2\\cdot\\simplify[unitFactor]{{a}x^3}-x^4 \\, dx =\\\\[3mm]
\\frac{\\simplify{{R0*b^2*c}}}{\\simplify{{6*a^3}}M}\\left[\\simplify[unitFactor]{{a^4}x} - \\simplify[unitFactor]{{a^3}x^2} +\\frac{\\simplify[unitFactor]{{a}x^4}}{2}-\\frac{x^5}{5} \\right]_0^\\var{a} = \\\\[3mm]

\\simplify[fractionNumbers]{({a}^5 - ({a}^5) + {a}^5 /2- {a}^5/5) *{R0*b^2*c}/{6*a^3*Mass}} =
\\simplify[fractionNumbers,otherNumbers]{({a}^5 - ({a}^5) + {a}^5 /2- {a}^5/5) *{R0*b^2*c}/{6*a^3*Mass}}.\\]

Finaly, for $z$ coordinate:

\\[\\bar z= \\frac{1}{M}\\int_T z\\,dM = \\int_0^\\var{a}\\int_0^{\\frac{\\var{c}}{\\var{a}}(\\var{a} -x)}\\int_0^{\\var{b}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{z}{\\var{c}}\\right)}z\\rho(x)\\,dy\\,dz\\,dx =\\\\[3mm]
\\frac{1}{M}\\int_0^\\var{a}\\int_0^{\\frac{\\var{c}}{\\var{a}}(\\var{a} -x)}\\int_0^{\\var{b}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{z}{\\var{c}}\\right)} z\\var{R0}(\\var{a}+x) \\,dy\\,dz\\,dx =\\\\[3mm]
\\frac{1}{M} \\int_0^\\var{a}\\var{R0}(\\var{a}+x)\\int_0^{\\frac{\\var{c}}{\\var{a}}(\\var{a} -x)}z\\int_0^{\\var{b}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{z}{\\var{c}}\\right)} \\,dy\\,dz\\,dx =\\\\[3mm]
\\frac{1}{M}\\int_0^\\var{a} \\var{R0}(\\var{a}+x)\\int_0^{\\frac{\\var{c}}{\\var{a}}(\\var{a} -x)} z\\var{b}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{z}{\\var{c}}\\right)\\,dz\\,dx =\\\\[3mm]
\\frac{1}{M}\\int_0^\\var{a} \\var{R0}(\\var{a}+x)\\int_0^{\\frac{\\var{c}}{\\var{a}}(\\var{a} -x)} \\var{b}\\left( z - \\frac{xz}{\\var{a}} - \\frac{z^2}{\\var{c}}\\right)\\,dz\\,dx =\\\\[3mm]
\\frac{\\simplify{{R0*a}}}{M}\\int_0^\\var{a} (\\var{a}+x) \\left[ \\frac{z^2}{2} - \\frac{xz^2}{\\var{2*a}} - \\frac{z^3}{\\var{3*c}}\\right]_0^{\\frac{\\var{c}}{\\var{a}}(\\var{a} -x)}\\,dx =\\\\[3mm]
\\frac{\\simplify{{R0*c^2}}}{\\simplify{{6*a^2}}M}\\int_0^\\var{a} (\\var{a}+x)(a-x)^3\\,dx =
\\frac{\\simplify{{R0*c^2}}}{\\simplify{{6*a^2}}M}\\int_0^\\var{a} \\var{a^4}-\\var{2*a^3}x + \\var{2*a}x^3 - x^4\\,dx\\\\[3mm]
\\frac{\\simplify{{R0*c^2}}}{\\simplify{{6*a^2}}M}\\left[\\var{a^4}x -\\frac{\\var{2*a^3}x^2}{2} + \\frac{\\var{a}x^4}{2} - \\frac{x^5}{5}\\right]_0^{\\var{a}}=\\\\[3mm]

\\simplify[fractionNumbers]{({a^5} -{2*a^5}/2 + {a^5}/2 - {a^5}/5) *{R0*c^2}/{6*a^2*Mass}} =
\\simplify[fractionNumbers,otherNumbers]{({a^5} -{2*a^5}/2 + {a^5}/2 - {a^5}/5) *{R0*c^2}/{6*a^2*Mass}}.
\\]

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Find the mass of the tetrahedron.

(enter your answer as an interger or a fraction)

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Find the centre of mass of the tetrahedron:

$\\bar x$ = [[0]]

$\\bar y$ = [[1]]

$\\bar z$ = [[2]]

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Double integral in polar coordinates - concentric circles

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Compute the integral

\\[\\iint_R \\frac{y^2}{\\sqrt{x^2 + y^2}}\\, dx\\, dy\\]

where $R$ is the region between the circles $x^2 + y^2 = \\var{r1}$ and $x^2 + y^2 = \\var{r2}$.

", "advice": "

Recall that ew pass to polar coordinates using $x=r\\cos(\\theta)$ and $r\\sin(\\theta)$.

a) Plug the values above into $ \\frac{y^2}{\\sqrt{x^2 + y^2}}$ we get

\\[ \\frac{r^2\\sin^2(\\theta)}{\\sqrt{r^2\\cos^2(\\theta) + r^2\\sin^2(\\theta)}} = \\frac{r^2\\sin^2(\\theta)}{r\\sqrt{\\cos^2(\\theta) + \\sin^2(\\theta)}} = r\\sin^2(\\theta).\\]

b) The area we are interested is the area between the concentric circles below (this shape is called and annulus)

{diagram}

This are can be described by the inequalities

\\[0<\\theta\\leq 2\\pi \\quad \\mbox{ and }\\quad \\var{r1}<r<\\var{r2}\\]

in polar coordinates.

c) Note that the order the integration does not matter as boundaries for both variables are constants. The integral thus can be written as

\\[\\int_{\\var{r1}}^{\\var{r2}}\\int_{0}^{2\\pi} r^2\\sin^2(\\theta)\\, d\\theta\\,dr.\\]

\\[\\int_{\\var{r1}}^{\\var{r2}}\\int_{0}^{2\\pi} r^2\\sin^2(\\theta)\\, d\\theta\\,dr = \\int_{\\var{r1}}^{\\var{r2}}r^2 \\int_{0}^{2\\pi} \\sin^2(\\theta)\\, d\\theta\\,dr = -\\int_{\\var{r1}}^{\\var{r2}}r^2 \\left[\\frac{\\sin(2\\theta) - 2\\theta}{4}\\right]_0^{2\\pi}\\,dr
= \\\\[3mm ]\\int_{\\var{r1}}^{\\var{r2}}r^2 \\pi\\,dr = \\pi\\left[\\frac{r^3}{3}\\right]_{\\var{r1}}^{\\var{r2}} = \\pi\\left(\\frac{\\var{r2}^2 - \\var{r1}^2}{3}\\right) = \\var{ans}.\\]

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Rewrite the function $ \\frac{y^2}{\\sqrt{x^2 + y^2}}$ in polar coordinates

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Parametrise the region $R$ in polar coordinate

[[0]]$<\\theta<$ [[1]] and [[2]] $<r<$[[3]]

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Write down the integral in polar coordinates

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Evaluate the integral you wrote in the previous part:

[[0]]

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moment of inertia - cylinder

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Determine the moment of inertia of the cylinder $C$ of uniform density $\\rho = \\var{d}$, height $\\var{h}$, and radius $\\var{a}$ about the axis passing through the centres of its circular faces.

", "advice": "

We can, naturally, consider this problem in cylindricla coordinates.

We may assume that the cylinder $C$ sits in the space such that the axis rotation described in the question is the $z$-axis.

Then $C$ can be described by

\\[0\\leq r \\leq \\var{a} \\quad \\mbox{and } \\quad 0\\leq \\theta\\leq 2\\pi \\quad\\mbox{and}\\quad 0\\leq z\\leq h.\\]

In cylindrical coordinates, the mass element is $dM = \\rho dV = \\rho r\\,d\\theta\\,dr\\, dz$. Also observe that each mass element is at the distance of $r$ to the axis of rotation. Then the moment of inertia is

\\[I_z = \\int_C r^2 \\, dM = \\int_C r^2\\rho\\, dV =\\iiint_C r^3\\rho\\, d\\theta\\,dr\\, dz =\\\\[3mm] \\int_0^{\\var{h}}\\int_0^{2\\pi}\\int_0^{\\var{a}} r^3\\rho\\, dr\\,d\\theta \\, dz = \\rho\\int_0^{\\var{h}}\\, dz \\int_0^{2\\pi}\\,d\\theta \\int_0^{\\var{a}} r^3\\, dr . \\]

Note that for the last equality we used the fact that boundaries of the integrals are constants.

Thus we obtain

\\[I_z = \\rho [z]_0^{\\var{h}} [\\theta]_0^{2\\pi}\\left[ \\frac{r^4}{4}\\right]_0^{\\var{a}} = \\frac{1}{2}\\pi\\rho\\times\\var{h}\\times{a^4} = \\simplify{1/2*pi*{d}*{h}*{a}^4}.\\]

Also note that $I_z = \\frac{1}{2}M\\var{a}^2$ where $M = \\rho V$, with $V=\\pi\\times\\var{a}^2\\times\\var{h}$.

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moment of inertia - cylinder

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Determine the moment of inertia of the cylinder $C$ of density varies as $\\rho = \\var{d}xy$, height $\\var{h}$, and radius $\\var{a}$ about the axis passing through the centres of its circular faces.

", "advice": "

We can, naturally, consider this problem in cylindricla coordinates.

We may assume that the cylinder $C$ sits in the space such that the axis rotation described in the question is the $z$-axis.

Then $C$ can be described by

\\[0\\leq r \\leq \\var{a} \\quad \\mbox{and } \\quad 0\\leq \\theta\\leq 2\\pi \\quad\\mbox{and}\\quad 0\\leq z\\leq h.\\]

We also have to turn $\\rho(x,y)$ into polar coordinates.

\\[\\rho(x,y) = \\var{d}(r\\cos\\theta)(r\\sin\\theta) = \\var{d}\\cdot r^2\\cos\\theta\\sin\\theta.\\]

Then, in cylindrical coordinates, the mass element becomes $dM = \\rho dV = r^2\\cos\\theta\\sin\\theta \\,r\\,d\\theta\\,dr\\, dz$. Also, observe that each mass element is at the distance of $r$ to the axis of rotation. Then the moment of inertia is

\\[I_z = \\int_C r^2 \\, dM = \\int_C r^2\\rho\\, dV =\\iiint_C \\var{d}\\cdot r^5 \\cos\\theta\\sin\\theta\\, d\\theta\\,dr\\, dz =\\\\[3mm] \\var{d}\\int_0^{\\var{h}}\\int_0^{2\\pi}\\int_0^{\\var{a}} r^5\\cos\\theta\\sin\\theta\\, dr\\,d\\theta \\, dz = \\var{d}\\int_0^{\\var{h}}\\, dz \\int_0^{2\\pi} \\cos\\theta\\sin\\theta\\,d\\theta \\int_0^{\\var{a}} r^5\\, dr . \\]

Note that for the last equality we used the fact that boundaries of the integrals are constants.

We can compute the integrals separately as follows:

\\[\\int_0^{\\var{h}}\\, dz = [z]_0^{\\var{h}} = \\var{h}\\]
\\[\\int_0^{2\\pi} \\cos\\theta\\sin\\theta\\,d\\theta = \\left[-\\frac{\\cos^2\\theta}{2}\\right]_0^{2\\pi} = \\frac{1-\\cos^2(2\\pi)}{2} =0.\\]
\\[\\int_0^{\\var{a}} r^5\\, dr = \\left[\\frac{r^6}{6}\\right]_0^{\\var{a}} = \\frac{\\var{a^6}}{6}.\\]

Thus we obtain

\\[I_z = \\var{d}\\times\\var{h}\\times 0\\times \\frac{\\var{a^6}}{6} = 0.\\]

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