{ "hash": "3b26136e4698c2a7035a4400b494be3b", "result": { "markdown": "# Surface Area\n\n\n\nThis section uses these add-on packages:\n\n``` {.julia .cell-code}\nusing CalculusWithJulia\nusing Plots\nusing SymPy\nusing QuadGK\n```\n\n\n\n\n---\n\n\n## Surfaces of revolution\n\n![ The exterior of the Jimi Hendrix Museum in Seattle has the signature style of its architect Frank Gehry. The surface is comprised of patches. A general method to find the amount of material to cover the surface - the surface area - might be to add up the area of *each* of the patches. However, in this section we will see for surfaces of revolution, there is an easier way. (Photo credit to [firepanjewellery](http://firepanjewellery.com/).) ](../integrals/figures/gehry-hendrix.jpg)\n\n\n\n> The surface area generated by rotating the graph of $f(x)$ between $a$ and $b$ about the $x$-axis is given by the integral\n>\n> $$\n> \\int_a^b 2\\pi f(x) \\cdot \\sqrt{1 + f'(x)^2} dx.\n> $$\n>\n> If the curve is parameterized by $(g(t), f(t))$ between $a$ and $b$ then the surface area is\n>\n> $$\n> \\int_a^b 2\\pi f(t) \\cdot \\sqrt{g'(t)^2 + f'(t)^2} dx.\n> $$\n>\n> These formulas do not add in the surface area of either of the ends.\n\n::: {.cell hold='true' execution_count=5}\n\n::: {.cell-output .cell-output-display execution_count=6}\n![](surface_area_files/figure-html/cell-6-output-1.svg){}\n:::\n:::\n\n\nThe above figure shows a cone (the line $y=x$) presented as a surface of revolution about the $x$-axis.\n\n\nTo see why this formula is as it is, we look at the parameterized case, the first one being a special instance with $g(t) =t$.\n\n\nLet a partition of $[a,b]$ be given by $a = t_0 < t_1 < t_2 < \\cdots < t_n =b$. This breaks the curve into a collection of line segments. Consider the line segment connecting $(g(t_{i-1}), f(t_{i-1}))$ to $(g(t_i), f(t_i))$. Rotating this around the $x$ axis will generate something approximating a disc, but in reality will be the frustum of a cone. What will be the surface area?\n\n\nConsider a right-circular cone parameterized by an angle $\\theta$ and the largest radius $r$ (so that the height satisfies $r/h=\\tan(\\theta)$). If this cone were made of paper, cut up a side, and layed out flat, it would form a sector of a circle, whose area would be $R\\gamma$ where $R$ is the radius of the circle (also the side length of our cone), and $\\gamma$ an angle that we can figure out from $r$ and $\\theta$. To do this, we note that the arc length of the circle's edge is $R\\gamma$ and also the circumference of the bottom of the cone so $R\\gamma = 2\\pi r$. With all this, we can solve to get $A = \\pi r^2/\\sin(\\theta)$. But we have a frustum of a cone with radii $r_0$ and $r_1$, so the surface area is a difference: $A = \\pi (r_1^2 - r_0^2) /\\sin(\\theta)$.\n\n\nRelating this to our values in terms of $f$ and $g$, we have $r_1=f(t_i)$, $r_0 = f(t_{i-1})$, and $\\sin(\\theta) = \\Delta f / \\sqrt{(\\Delta g)^2 + (\\Delta f)^2}$, where $\\Delta f = f(t_i) - f(t_{i-1})$ and similarly for $\\Delta g$.\n\n\nPutting this altogether we get that the surface area generarated by rotating the line segment around the $x$ axis is\n\n\n\n$$\n\\text{sa}_i = \\pi (f(t_i)^2 - f(t_{i-1})^2) \\cdot \\sqrt{(\\Delta g)^2 + (\\Delta f)^2} / \\Delta f =\n\\pi (f(t_i) + f(t_{i-1})) \\cdot \\sqrt{(\\Delta g)^2 + (\\Delta f)^2}.\n$$\n\n\n(This is $2 \\pi$ times the average radius times the slant height.)\n\n\nAs was done in the derivation of the formula for arc length, these pieces are multiplied both top and bottom by $\\Delta t = t_{i} - t_{i-1}$. Carrying the bottom inside the square root and noting that by the mean value theorem $\\Delta g/\\Delta t = g(\\xi)$ and $\\Delta f/\\Delta t = f(\\psi)$ for some $\\xi$ and $\\psi$ in $[t_{i-1}, t_i]$, this becomes:\n\n\n\n$$\n\\text{sa}_i = \\pi (f(t_i) + f(t_{i-1})) \\cdot \\sqrt{(g'(\\xi))^2 + (f'(\\psi))^2} \\cdot (t_i - t_{i-1}).\n$$\n\n\nAdding these up, $\\text{sa}_1 + \\text{sa}_2 + \\cdots + \\text{sa}_n$, we get a Riemann sum approximation to the integral\n\n\n\n$$\n\\text{SA} = \\int_a^b 2\\pi f(t) \\sqrt{g'(t)^2 + f'(t)^2} dt.\n$$\n\n\nIf we assume integrability of the integrand, then as our partition size goes to zero, this approximate surface area converges to the value given by the limit. (As with arc length, this needs a technical adjustment to the Riemann integral theorem as here we are evaluating the integrand function at four points ($t_i$, $t_{i-1}$, $\\xi$ and $\\psi$) and not just at some $c_i$.\n\n::: {.cell cache='true' hold='true' execution_count=6}\n\n::: {.cell-output .cell-output-display execution_count=7}\n```{=html}\n
\"A\n

Surface of revolution of \\(f(x) = 2 - x^2\\) about the \\(y\\) axis. The lines segments are the images of rotating the secant line connecting \\((1/2, f(1/2))\\) and \\((3/4, f(3/4))\\). These trace out the frustum of a cone which approximates the corresponding surface area of the surface of revolution. In the limit, this approximation becomes exact and a formula for the surface area of surfaces of revolution can be used to compute the value.

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\n```\n:::\n:::\n\n\n#### Examples\n\n\nLets see that the surface area of an open cone follows from this formula, even though we just saw how to get this value.\n\n\nA cone be be envisioned as rotating the function $f(x) = x\\tan(\\theta)$ between $0$ and $h$ around the $x$ axis. This integral yields the surface area:\n\n\n\n$$\n\\begin{align*}\n\\int_0^h 2\\pi f(x) \\sqrt{1 + f'(x)^2}dx\n&= \\int_0^h 2\\pi x \\tan(\\theta) \\sqrt{1 + \\tan(\\theta)^2}dx \\\\\n&= (2\\pi\\tan(\\theta)\\sqrt{1 + \\tan(\\theta)^2} x^2/2 \\big|_0^h \\\\\n&= \\pi \\tan(\\theta) \\sec(\\theta) h^2 \\\\\n&= \\pi r^2 / \\sin(\\theta).\n\\end{align*}\n$$\n\n\n(There are many ways to express this, we used $r$ and $\\theta$ to match the work above. If the cone is parameterized by a height $h$ and radius $r$, then the surface area of the sides is $\\pi r\\sqrt{h^2 + r^2}$. If the base is included, there is an additional $\\pi r^2$ term.)\n\n\n##### Example\n\n\nLet the graph of $f(x) = x^2$ from $x=0$ to $x=1$ be rotated around the $x$ axis. What is the resulting surface area generated?\n\n\n\n$$\n\\text{SA} = \\int_a^b 2\\pi f(x) \\sqrt{1 + f'(x)^2}dx = \\int_0^1 2\\pi x^2 \\sqrt{1 + (2x)^2} dx.\n$$\n\n\nThis integral is done by a trig substitution, but gets involved. We let `SymPy` do it:\n\n::: {.cell execution_count=7}\n``` {.julia .cell-code}\n@syms x\nF = integrate(2 * PI * x^2 * sqrt(1 + (2x)^2), x)\n```\n\n::: {.cell-output .cell-output-display execution_count=8}\n```{=html}\n \n\\[\n2 \\pi \\left(\\frac{x^{5}}{\\sqrt{4 x^{2} + 1}} + \\frac{3 x^{3}}{8 \\sqrt{4 x^{2} + 1}} + \\frac{x}{32 \\sqrt{4 x^{2} + 1}} - \\frac{\\operatorname{asinh}{\\left(2 x \\right)}}{64}\\right)\n\\]\n\n```\n:::\n:::\n\n\nWe show `F`, only to demonstrate that indeed the integral is a bit involved. The actual surface area follows from a *definite* integral, which we get through the fundamental theorem of calculus:\n\n::: {.cell execution_count=8}\n``` {.julia .cell-code}\nF(1) - F(0)\n```\n\n::: {.cell-output .cell-output-display execution_count=9}\n```{=html}\n \n\\[\n2 \\pi \\left(- \\frac{\\operatorname{asinh}{\\left(2 \\right)}}{64} + \\frac{9 \\sqrt{5}}{32}\\right)\n\\]\n\n```\n:::\n:::\n\n\n### Plotting surfaces of revolution\n\n\nThe commands to plot a surface of revolution will be described more clearly later; for now we present them as simply a pattern to be followed in case plots are desired. Suppose the curve in the $x-y$ plane is given parametrically by $(g(u), f(u))$ for $a \\leq u \\leq b$.\n\n\nTo be concrete, we parameterize the circle centered at $(6,0)$ with radius $2$ by:\n\n::: {.cell execution_count=9}\n``` {.julia .cell-code}\ng(u) = 6 + 2sin(u)\nf(u) = 2cos(u)\na, b = 0, 2pi\n```\n\n::: {.cell-output .cell-output-display execution_count=10}\n```\n(0, 6.283185307179586)\n```\n:::\n:::\n\n\nThe plot of this curve is:\n\n::: {.cell hold='true' execution_count=10}\n``` {.julia .cell-code}\nus = range(a, b, length=100)\nplot(g.(us), f.(us), xlims=(-0.5, 9), aspect_ratio=:equal, legend=false)\nplot!([0,0],[-3,3], color=:red, linewidth=5) # y axis emphasis\nplot!([3,9], [0,0], color=:green, linewidth=5) # x axis emphasis\n```\n\n::: {.cell-output .cell-output-display execution_count=11}\n![](surface_area_files/figure-html/cell-11-output-1.svg){}\n:::\n:::\n\n\nThough parametric plots have a convenience constructor, `plot(g, f, a, b)`, we constructed the points with `Julia`'s broadcasting notation, as we will need to do for a surface of revolution. The `xlims` are adjusted to show the $y$ axis, which is emphasized with a layered line. The line is drawn by specifying two points, $(x_0, y_0)$ and $(x_1, y_1)$ in the form `[x0,x1]` and `[y0,y1]`.\n\n\nNow, to rotate this about the $y$ axis, creating a surface plot, we have the following pattern:\n\n::: {.cell execution_count=11}\n``` {.julia .cell-code}\nS(u,v) = [g(u)*cos(v), g(u)*sin(v), f(u)]\nus = range(a, b, length=100)\nvs = range(0, 2pi, length=100)\nws = unzip(S.(us, vs')) # reorganize data\nsurface(ws..., zlims=(-6,6), legend=false)\n\nplot!([0,0], [0,0], [-3,3], color=:red, linewidth=5) # y axis emphasis\n```\n\n::: {.cell-output .cell-output-display execution_count=12}\n![](surface_area_files/figure-html/cell-12-output-1.svg){}\n:::\n:::\n\n\nThe `unzip` function is not part of base `Julia`, rather part of `CalculusWithJulia`. This function rearranges data into a form consumable by the plotting methods like `surface`. In this case, the result of `S.(us,vs')` is a grid (matrix) of points, the result of `unzip` is three grids of values, one for the $x$ values, one for the $y$ values, and one for the $z$ values. A manual adjustment to the `zlims` is used, as `aspect_ratio` does not have an effect with the `plotly()` backend and errors on 3d graphics with `pyplot()`.\n\n\nTo rotate this about the $x$ axis, we have this pattern:\n\n::: {.cell hold='true' execution_count=12}\n``` {.julia .cell-code}\nS(u,v) = [g(u), f(u)*cos(v), f(u)*sin(v)]\nus = range(a, b, length=100)\nvs = range(0, 2pi, length=100)\nws = unzip(S.(us,vs'))\nsurface(ws..., legend=false)\n\nplot!([3,9], [0,0],[0,0], color=:green, linewidth=5) # x axis emphasis\n```\n\n::: {.cell-output .cell-output-display execution_count=13}\n![](surface_area_files/figure-html/cell-13-output-1.svg){}\n:::\n:::\n\n\nThe above pattern covers the case of rotating the graph of a function $f(x)$ of $a,b$ by taking $g(t)=t$.\n\n\n##### Example\n\n\nRotate the graph of $x^x$ from $0$ to $3/2$ around the $x$ axis. What is the surface area generated?\n\n\nWe work numerically for this one, as no antiderivative is forthcoming. Recall, the accompanying `CalculusWithJulia` package defines `f'` to return the automatic derivative through the `ForwardDiff` package.\n\n::: {.cell hold='true' execution_count=13}\n``` {.julia .cell-code}\nf(x) = x^x\na, b = 0, 3/2\nval, _ = quadgk(x -> 2pi * f(x) * sqrt(1 + f'(x)^2), a, b)\nval\n```\n\n::: {.cell-output .cell-output-display execution_count=14}\n```\n14.934256764843937\n```\n:::\n:::\n\n\n(The function is not defined at $x=0$ mathematically, but is on the computer to be $1$, the limiting value. Even were this not the case, the `quadgk` function doesn't evaluate the function at the points `a` and `b` that are specified.)\n\n::: {.cell hold='true' execution_count=14}\n``` {.julia .cell-code}\ng(u) = u\nf(u) = u^u\nS(u,v) = [g(u)*cos(v), g(u)*sin(v), f(u)]\nus = range(0, 3/2, length=100)\nvs = range(0, pi, length=100) # not 2pi (to see inside)\nws = unzip(S.(us,vs'))\nsurface(ws..., alpha=0.75)\n```\n\n::: {.cell-output .cell-output-display execution_count=15}\n![](surface_area_files/figure-html/cell-15-output-1.svg){}\n:::\n:::\n\n\nWe compare this answer to that of the frustum of a cone with radii $1$ and $(3/2)^2$, formed by rotating the line segment connecting $(0,f(0))$ with $(3/2,f(3/2))$. From looking at the graph of the surface, these values should be comparable. The surface area of the cone part is $\\pi (r_1^2 + r_0^2) / \\sin(\\theta) = \\pi (r_1 + r_0) \\cdot \\sqrt{(\\Delta h)^2 + (r_1-r_0)^2}$.\n\n::: {.cell hold='true' execution_count=15}\n``` {.julia .cell-code}\nf(x) = x^x\nr0, r1 = f(0), f(3/2)\npi * (r1 + r0) * sqrt((3/2)^2 + (r1-r0)^2)\n```\n\n::: {.cell-output .cell-output-display execution_count=16}\n```\n15.310680925915081\n```\n:::\n:::\n\n\n##### Example\n\n\nWhat is the surface area generated by Gabriel's Horn, the solid formed by rotating $1/x$ for $x \\geq 1$ around the $x$ axis?\n\n\n\n$$\n\\text{SA} = \\int_a^b 2\\pi f(x) \\sqrt{1 + f'(x)^2}dx =\n\\lim_{M \\rightarrow \\infty} \\int_1^M 2\\pi \\frac{1}{x} \\sqrt{1 + (-1/x^2)^2} dx.\n$$\n\n\nWe do this with `SymPy`:\n\n::: {.cell execution_count=16}\n``` {.julia .cell-code}\n@syms M\nex = integrate(2PI * (1/x) * sqrt(1 + (-1/x)^2), (x, 1, M))\n```\n\n::: {.cell-output .cell-output-display execution_count=17}\n```{=html}\n \n\\[\n2 \\pi \\left(- \\frac{M}{\\sqrt{M^{2} + 1}} + \\operatorname{asinh}{\\left(M \\right)} - \\frac{1}{M \\sqrt{M^{2} + 1}}\\right) - 2 \\pi \\left(- \\sqrt{2} + \\log{\\left(1 + \\sqrt{2} \\right)}\\right)\n\\]\n\n```\n:::\n:::\n\n\nThe limit as $M$ gets large is of interest. The only term that might get out of hand is `asinh(M)`. We check its limit:\n\n::: {.cell execution_count=17}\n``` {.julia .cell-code}\nlimit(asinh(M), M => oo)\n```\n\n::: {.cell-output .cell-output-display execution_count=18}\n```{=html}\n \n\\[\n\\infty\n\\]\n\n```\n:::\n:::\n\n\nSo indeed it does. There is nothing to balance this out, so the integral will be infinite, as this shows:\n\n::: {.cell execution_count=18}\n``` {.julia .cell-code}\nlimit(ex, M => oo)\n```\n\n::: {.cell-output .cell-output-display execution_count=19}\n```{=html}\n \n\\[\n\\infty\n\\]\n\n```\n:::\n:::\n\n\nThis figure would have infinite surface, were it possible to actually construct an infinitely long solid. (But it has been shown to have *finite* volume.)\n\n\n##### Example\n\n\nThe curve described parametrically by $g(t) = 2(1 + \\cos(t))\\cos(t)$ and $f(t) = 2(1 + \\cos(t))\\sin(t)$ from $0$ to $\\pi$ is rotated about the $x$ axis. Find the resulting surface area.\n\n\nThe graph shows half a heart, the resulting area will resemble an apple.\n\n::: {.cell hold='true' execution_count=19}\n``` {.julia .cell-code}\ng(t) = 2(1 + cos(t)) * cos(t)\nf(t) = 2(1 + cos(t)) * sin(t)\nplot(g, f, 0, 1pi)\n```\n\n::: {.cell-output .cell-output-display execution_count=20}\n![](surface_area_files/figure-html/cell-20-output-1.svg){}\n:::\n:::\n\n\nThe integrand simplifies to $8\\sqrt{2}\\pi \\sin(t) (1 + \\cos(t))^{3/2}$. This lends itself to $u$-substitution with $u=\\cos(t)$.\n\n\n\n$$\n\\begin{align*}\n\\int_0^\\pi 8\\sqrt{2}\\pi \\sin(t) (1 + \\cos(t))^{3/2}\n&= 8\\sqrt{2}\\pi \\int_1^{-1} (1 + u)^{3/2} (-1) du\\\\\n&= 8\\sqrt{2}\\pi (2/5) (1+u)^{5/2} \\big|_{-1}^1\\\\\n&= 8\\sqrt{2}\\pi (2/5) 2^{5/2} = \\frac{2^7 \\pi}{5}.\n\\end{align*}\n$$\n\n\n## The first Theorem of Pappus\n\n\nThe [first](http://tinyurl.com/le3lvb9) theorem of Pappus provides a simpler means to compute the surface area if the distance the centroid is from the axis ($\\rho$) and the arc length of the curve ($L$) are both known. In that case, the surface area satisfies:\n\n\n\n$$\n\\text{SA} = 2 \\pi \\rho L\n$$\n\n\nThat is, the surface area is simply the circumference of the circle traced out by the centroid of the curve times the length of the curve - the distances rotated are collapsed to that of just the centroid.\n\n\n##### Example\n\n\nThe surface area of of an open cone can be computed, as the arc length is $\\sqrt{h^2 + r^2}$ and the centroid of the line is a distance $r/2$ from the axis. This gives SA$=2\\pi (r/2) \\sqrt{h^2 + r^2} = \\pi r \\sqrt{h^2 + r^2}$.\n\n\n##### Example\n\n\nWe can get the surface area of a torus from this formula.\n\n\nThe torus is found by rotating the curve $(x-b)^2 + y^2 = a^2$ about the $y$ axis. The centroid is $b$, the arc length $2\\pi a$, so the surface area is $2\\pi (b) (2\\pi a) = 4\\pi^2 a b$.\n\n\nA torus with $a=2$ and $b=6$\n\n::: {.cell hold='true' execution_count=20}\n\n::: {.cell-output .cell-output-display execution_count=21}\n![](surface_area_files/figure-html/cell-21-output-1.svg){}\n:::\n:::\n\n\n##### Example\n\n\nThe surface area of sphere will be SA$=2\\pi \\rho (\\pi r) = 2 \\pi^2 r \\cdot \\rho$. What is $\\rho$? The centroid of an arc formula can be derived in a manner similar to that of the centroid of a region. The formulas are:\n\n\n\n$$\n\\begin{align}\n\\text{cm}_x &= \\frac{1}{L} \\int_a^b g(t) \\sqrt{g'(t)^2 + f'(t)^2} dt\\\\\n\\text{cm}_y &= \\frac{1}{L} \\int_a^b f(t) \\sqrt{g'(t)^2 + f'(t)^2} dt.\n\\end{align}\n$$\n\n\nHere, $L$ is the arc length of the curve.\n\n\nFor the sphere parameterized by $g(t) = r \\cos(t)$, $f(t) = r\\sin(t)$, we get that these become\n\n\n\n$$\n\\text{cm}_x = \\frac{1}{L}\\int_0^\\pi r\\cos(t) \\sqrt{r^2(\\sin(t)^2 + \\cos(t)^2)} dt = \\frac{1}{L}r^2 \\int_0^\\pi \\cos(t) = 0.\n$$\n\n$$\n\\text{cm}_y = \\frac{1}{L}\\int_0^\\pi r\\sin(t) \\sqrt{r^2(\\sin(t)^2 + \\cos(t)^2)} dt = \\frac{1}{L}r^2 \\int_0^\\pi \\sin(t) = \\frac{1}{\\pi r} r^2 \\cdot 2 = \\frac{2r}{\\pi}.\n$$\n\n\nCombining this, we see that the surface area of a sphere is $2 \\pi^2 r (2r/\\pi) = 4\\pi r^2$, by Pappus' Theorem.\n\n\n## Questions\n\n\n##### Questions\n\n\nThe graph of $f(x) = \\sin(x)$ from $0$ to $\\pi$ is rotated around the $x$ axis. After a $u$-substitution, what integral would give the surface area generated?\n\n::: {.cell hold='true' execution_count=21}\n\n::: {.cell-output .cell-output-display execution_count=22}\n```{=html}\n
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\n\n\n```\n:::\n:::\n\n\nThough the integral can be computed by hand, give a numeric value.\n\n::: {.cell hold='true' execution_count=22}\n\n::: {.cell-output .cell-output-display execution_count=23}\n```{=html}\n
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\n\n\n```\n:::\n:::\n\n\n##### Questions\n\n\nThe graph of $f(x) = \\sqrt{x}$ from $0$ to $4$ is rotated around the $x$ axis. Numerically find the surface area generated?\n\n::: {.cell hold='true' execution_count=23}\n\n::: {.cell-output .cell-output-display execution_count=24}\n```{=html}\n
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\n\n\n```\n:::\n:::\n\n\n##### Questions\n\n\nFind the surface area generated by revolving the graph of the function $f(x) = x^3/9$ from $x=0$ to $x=2$ around the $x$ axis. This can be done by hand or numerically.\n\n::: {.cell hold='true' execution_count=24}\n\n::: {.cell-output .cell-output-display execution_count=25}\n```{=html}\n
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\n\n\n```\n:::\n:::\n\n\n##### Questions\n\n\n(From Stewart.) If a loaf of bread is in the form of a sphere of radius $1$, the amount of crust for a slice depends on the width, but not where in the loaf it is sliced.\n\n\nThat is this integral with $f(x) = \\sqrt{1 - x^2}$ and $u, u+h$ in $[-1,1]$ does not depend on $u$:\n\n\n\n$$\nA = \\int_u^{u+h} 2\\pi f(x) \\sqrt{1 + f'(x)^2} dx.\n$$\n\n\nIf we let $f(x) = y$ then $f'(x) = x/y$. With this, what does the integral above come down to after cancellations:\n\n::: {.cell hold='true' execution_count=25}\n\n::: {.cell-output .cell-output-display execution_count=26}\n```{=html}\n
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\n\n\n```\n:::\n:::\n\n\n##### Questions\n\n\nFind the surface area of the dome of sphere generated by rotating the the curve generated by $g(t) = \\cos(t)$ and $f(t) = \\sin(t)$ for $t$ in $0$ to $\\pi/6$.\n\n\nNumerically find the value.\n\n::: {.cell hold='true' execution_count=26}\n\n::: {.cell-output .cell-output-display execution_count=27}\n```{=html}\n
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\n\n\n```\n:::\n:::\n\n\n##### Questions\n\n\nThe [astroid](http://www-history.mcs.st-and.ac.uk/Curves/Astroid.html) is parameterized by $g(t) = a\\cos(t)^3$ and $f(t) = a \\sin(t)^3$. Let $a=1$ and rotate the curve from $t=0$ to $t=\\pi$ around the $x$ axis. What is the surface area generated?\n\n::: {.cell hold='true' execution_count=27}\n\n::: {.cell-output .cell-output-display execution_count=28}\n```{=html}\n
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\n\n\n```\n:::\n:::\n\n\n##### Questions\n\n\nFor the curve parameterized by $g(t) = a\\cos(t)^5$ and $f(t) = a \\sin(t)^5$. Let $a=1$ and rotate the curve from $t=0$ to $t=\\pi$ around the $x$ axis. Numerically find the surface area generated?\n\n::: {.cell hold='true' execution_count=28}\n\n::: {.cell-output .cell-output-display execution_count=29}\n```{=html}\n
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\n\n\n```\n:::\n:::\n\n\n", "supporting": [ "surface_area_files" ], "filters": [], "includes": { "include-in-header": [ "\n\n\n" ] } } }