 # Can someone put up a tutorial on Integral Calculus?

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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 $A=\int_a^b dA$. This is equal to the double integral over the infinitesimal length $\int_a^b \int_c^d ds$ because $\int_c^d ds = dA$.

In 3 dimensions, ds is given by

$ds = \sqrt{dx^2+dy^2+dz^2}$

which is equivalent to

$\sqrt {1+\left(\frac{dz}{dx}\right)^2+\left(\frac{dz}{dy}\right)^2} dxdy$

Inserting the value of ds gives us the equation for the surface area of a function z=f(x, y):

$A=\int_a^b \int_c^d \sqrt{1+\left(\frac{dz}{dx}\right)^2+\left(\frac{dz}{dy}\right)^2}dxdy$

This can be generalized to n dimensions by saying

$S=\int_a^b \int_c^d \int_e^f... \sqrt{1+\left(\frac{df}{dx}\right)^2+\left(\frac{df}{dy}\right)^2+\left(\frac{df}{dz}\right)^2...\left(\frac{df}{dx_n}\right)^2} dxdydz...dx_n$

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 $f(x, y)=\sqrt x + \sqrt y$

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

$\int_6^{12} \int_4^7 \sqrt{1+x+y} dx dy$

The second antiderivative is

$\frac{4(x+y+1)^{\frac{5}{2}}}{15}$

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.

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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 $x=g(t)$ and $y=h(t)$. The parametrized curve will be written as a vector function $r(t)=g(t)i + h(t)j$. It is also assumed to be smooth. When a curve is smooth, $r(t)$ 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

$\int_C f(x, y) \ ds = \oint f(x, y) \ ds$

The ds means that the integral is with respect to the arc length, not x or y.

In parametric form,

$ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt$

Rewriting the equation in parametric form gives us:

$\oint f(x, y) \ ds = \int_a^b f(g(t), h(t)) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt$

Because

$\sqrt{\left(\frac{dx}{dt}\right) ^2 + \left(\frac{dy}{dt}\right) ^2}=\left|\left|\frac{dr}{dt}\right|\right|$

we obtain the equation for the line integral:

$\oint f(x, y) \ ds = \int_a^b f(g(t), h(t))\left|\left|\frac{dr}{dt}\right|\right|dt$

For example, we will work out the line integral from 2 to 6 of f(x, y) = x2 + y2, where x = cos(t) and y = sin(t). For this integral,

$\left|\left|\frac{dr}{dt}\right|\right|=\sqrt{1-2sin^2(t)}$

and

$f(g(t), h(t))=sin^2(t)+cos^2(t)=1$

The integral now becomes

$\int_2^6 \sqrt{1-sin^2(t)} \ dt$

The antiderivative of this function is

$\frac{tan(t)}{\sqrt{tan^2(t)+1}}$
Evaluating the antiderivative gives us

$\int_2^6 \sqrt{1-sin^2(t)} \ dt=0.63$

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:

$\oint f(x, y) \ ds = \sum \int_C f(x, y) \left|\left|\frac{dr}{dt}\right|\right| dt$

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

$dx=\frac{dx}{dt} dt$

and

$dy=\frac{dy}{dt}dt$

We can definite the line integral with respect to x or y as

$\oint f(x, y) dx = \int_a^b f(x(t), y(t))\frac{dx][dt}dt$

and

$\oint f(x, y) dy=\int_c^d f(x(t), y(t)) \frac{dy}{dt}dt$

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

$\int_6^9 cos(t) dt$

The antiderivative of the function is -sin(t).

Evaluating this gives us

$\int_6^9 cos(t) dt=-0.69$

That concludes the tutorial on integral calculus.

What will happen to these tutorials?

Edited by Endercreeper01

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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/

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