Endercreeper01 47 Posted January 11, 2014 Share Posted January 11, 2014 This is the last tutorial on integral calculus. If I forgot to do a tutorial on anything in integral calculus, remind me, but if not, this will be the final integral calculus tutorial. I will also provide practice problems upon request. If anyone wants more tutorials, I will make them. I have lots of time for tutorials, so time will not be a problem. Part 15: Surface area/ Surface volume of functions in n dimensions We will start off in 3 dimensions just for simplicity. To find the surface area, we have to integrate each infinitesimal area over the entire region, i.e [latex]A=\int_a^b dA[/latex]. This is equal to the double integral over the infinitesimal length [latex]\int_a^b \int_c^d ds[/latex] because [latex]\int_c^d ds = dA[/latex]. In 3 dimensions, ds is given by [latex]ds = \sqrt{dx^2+dy^2+dz^2}[/latex] which is equivalent to [latex]\sqrt {1+\left(\frac{dz}{dx}\right)^2+\left(\frac{dz}{dy}\right)^2} dxdy[/latex] Inserting the value of ds gives us the equation for the surface area of a function z=f(x, y): [latex]A=\int_a^b \int_c^d \sqrt{1+\left(\frac{dz}{dx}\right)^2+\left(\frac{dz}{dy}\right)^2}dxdy[/latex] This can be generalized to n dimensions by saying [latex]S=\int_a^b \int_c^d \int_e^f... \sqrt{1+\left(\frac{df}{dx}\right)^2+\left(\frac{df}{dy}\right)^2+\left(\frac{df}{dz}\right)^2...\left(\frac{df}{dx_n}\right)^2} dxdydz...dx_n[/latex] Where S is the n surface area/volume of the function in n-1 dimensions. For example, we will work this integral out when a = 6, b = 12, c = 4, and d = 7 of the function [latex]f(x, y)=\sqrt x + \sqrt y[/latex] The square of the partial derivative of f with respect to x or y will be x when taken with respect to x and y when taken with respect to y. This now becomes [latex]\int_6^{12} \int_4^7 \sqrt{1+x+y} dx dy[/latex] The second antiderivative is [latex]\frac{4(x+y+1)^{\frac{5}{2}}}{15}[/latex] and evaluating this expression for the surface area gives us A = 370.011408631 That concludes the tutorial on integral calculus. Remember, if you need any help with this, you can just ask our calculus forum. 1 Link to post Share on other sites

Endercreeper01 47 Posted January 13, 2014 Share Posted January 13, 2014 (edited) The actual final tutorial will be on line integrals. Part 16. Line Integrals With line integrals, x and y are coordinates on a curve C. First, we must parametrize the curve, i.e [latex]x=g(t)[/latex] and [latex]y=h(t)[/latex]. The parametrized curve will be written as a vector function [latex]r(t)=g(t)i + h(t)j[/latex]. It is also assumed to be smooth. When a curve is smooth, [latex]r(t)[/latex] is continuous and never 0 for any value of t. When you have a function f(x, y) along a curve C, the line integral is [latex]\int_C f(x, y) \ ds = \oint f(x, y) \ ds[/latex] The ds means that the integral is with respect to the arc length, not x or y. In parametric form, [latex]ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt[/latex] Rewriting the equation in parametric form gives us: [latex]\oint f(x, y) \ ds = \int_a^b f(g(t), h(t)) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt[/latex] Because [latex]\sqrt{\left(\frac{dx}{dt}\right) ^2 + \left(\frac{dy}{dt}\right) ^2}=\left|\left|\frac{dr}{dt}\right|\right|[/latex] we obtain the equation for the line integral: [latex]\oint f(x, y) \ ds = \int_a^b f(g(t), h(t))\left|\left|\frac{dr}{dt}\right|\right|dt[/latex] For example, we will work out the line integral from 2 to 6 of f(x, y) = x^{2} + y^{2}, where x = cos(t) and y = sin(t). For this integral, [latex]\left|\left|\frac{dr}{dt}\right|\right|=\sqrt{1-2sin^2(t)}[/latex] and [latex]f(g(t), h(t))=sin^2(t)+cos^2(t)=1[/latex] The integral now becomes [latex]\int_2^6 \sqrt{1-sin^2(t)} \ dt[/latex] The antiderivative of this function is [latex]\frac{tan(t)}{\sqrt{tan^2(t)+1}}[/latex]Evaluating the antiderivative gives us [latex]\int_2^6 \sqrt{1-sin^2(t)} \ dt=0.63[/latex] We can also find line integrals of piecewise smooth curves. A piecewise smooth curve is a smooth curve made up of parts. The line integral for a piecewise smooth curve is: [latex]\oint f(x, y) \ ds = \sum \int_C f(x, y) \left|\left|\frac{dr}{dt}\right|\right| dt[/latex] i.e it is the sum of it's parts. Part 17. Line Integrals with Respect to x and y Line integrals can also be taken with respect to x or y. Because [latex]dx=\frac{dx}{dt} dt[/latex] and [latex]dy=\frac{dy}{dt}dt[/latex] We can definite the line integral with respect to x or y as [latex]\oint f(x, y) dx = \int_a^b f(x(t), y(t))\frac{dx][dt}dt[/latex] and [latex]\oint f(x, y) dy=\int_c^d f(x(t), y(t)) \frac{dy}{dt}dt[/latex] For example, we will work out the integral from 6 to 9 of f(x, y) = x^2 + y^2 where x=sin(t) and y=cos(t). f(x, y) becomes 1, leaving us with [latex]\int_6^9 cos(t) dt[/latex] The antiderivative of the function is -sin(t). Evaluating this gives us [latex]\int_6^9 cos(t) dt=-0.69[/latex] That concludes the tutorial on integral calculus. What will happen to these tutorials? Edited January 13, 2014 by Endercreeper01 1 Link to post Share on other sites

hypervalent_iodine 1365 Posted January 15, 2014 Share Posted January 15, 2014 Just FYI: I've pinned some old tutorials related to this subject in this sub-forum. They may be found here: http://www.scienceforums.net/topic/4108-calculus-i-lesson-1-a-background-to-differentation/ http://www.scienceforums.net/topic/4182-calculus-i-lesson-2-a-continuation-from-first-principles/ Edit: actually, I decided to move them to the tutorials forum: http://www.scienceforums.net/forum/88-mathematics-tutorials/ Link to post Share on other sites

rahul_26 0 Posted September 24, 2016 Share Posted September 24, 2016 Try Actucation and Khan academy Link to post Share on other sites

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