how do i change the domain in a double integral from rectangular coordinates to polar coordinates?
968004
To change from Cartesian coordinates (x,y) to Polar Coordinates (r, theta), you need to perform the following substitutions:\n\nx = r*cos(theta)\ny = r*sin(theta)\n\nand:\n\nr = sqrt(x^2+y^2)\ntheta = arctan(y/x)\n\nThe general form of the integral substitution is:\n\ndouble integral[f(x,y)dydx] = double integral[f(r*cos(theta), r*sin(theta))*r*dr*d(theta)]\n\nBelow is a website with some examples\n\n------------\n\nHope this helps
897617
This is an equation with separated variables. Rewrite as \n\ndu/u^3 = (t^3-t)dt \n\nand integrate (find antideritive) of each side separately.
6838
∫xSec(x²)²dx = .5Sin(x²)/Cos(x²)
719380
Let y = x^(1/2)\n\nso y^2 = x,\n\ndx = 2y dy\n\nThe problem becomes:\n\nint [1 / (y + y^3)] 2y dy\n\nOr:\n\nint 2 [1 / (1 + y^2)] dy\n\nA recognizable integral. Integrate and substitute back x^1/2 = y
323916
sec is 1/cos. So the domain is the set of all real numbers except when cos(2x - pi) is 0. To graph it, first plot the cosine graph corresponding to -2cos(2x - pi). Then, "flip" the wave along y=2 and y=-2 for when the graph is positive and negative. Basically, what you get are sets of parabolas going off to infinity alternating between being above and below the x axes. Then, erase your cosine graph and shift these "parabolas" up by 3, and you're good.\nSince -2cos can take any value between -2 and 2, -2sec can take any value except those between -2 and 2. Shift up by 3, and you get a range of all real numbers less than or equal to 1 and greater than or equal to 4.
390806
The above theorem about products of integrals is wrong as stated. Counterexample: Int[1]dx = Int[1*1]dx = x + C. But (Int[1]dx)*(Int[1]dx) = x^2 + ax + b.\nInstead, use trigonometric substitution to change square roots into trigonometric functions. Sqrt[25-4x^2] can be written as the leg of a right triangle with hypotenuse 5 and another leg of length 2x. If we let the angle adjacent to the leg of length Sqrt[25-4x^2] be t, then 5cos(t) is the main part of your integrand and x = (5/2)sin(t) implies dx = (5/2)cos(t) dt.
979191
When converting from rectangular coordinates to polar coordinated, remember the following.\n\n@ = Theta\nx = r*cos(@)\ny =r*sin(@)\n\nr^2 = x^2 +y^2\n\ntan(@) = y/x\n\nso your equation\nx^2 +y^2 =36 becomes\nr^2 = 36   OR\nr =6 (which is a circle with radius 6)
68076
god man.  what kind of horrible painful calculus class are you taking?\n\nfirst integrate over dx.  the integral of x^y dx is just a power integral, so you have y*x^(y-1).  the integral of y^x dx is just an exponential, so you have y^x/ln y.\n\nnow integrate y*x^(y-1) + y^x/ln y over dy.  and there, my friend, is where i stop because frankly i can't be arsed to do something so brain-curdling.  actually the first term doesn't look so bad, i think if you integrate it by parts it should come out in one step.  don't really want to think about doing the 2nd one.\n\nyou might consider trying some kind of fancy thing like transforming to polar coordinates or something.  might make it easier.\n\nbtw you should know about this site:\nhttp://integrals.wolfram.com/
631241
the integration by part is very long and complicated\ni used it many times\nlet u =logsinx          dv=dx\n   du= cosx/sinx          =x\nxlogsinx-int{xcosx/sinx}\nint{xcosx/sinx}  by sbutitution\nlet u=sinx   x=sin^-1(x)\ndu=cosx dx dx=du/cosx\nyou will get int{sin^-1(u)/u} remmeber u=sinx\nby part\nlet k=1/u          dv=sin^-1(u)\n     -1/u^2        int{sin^-1(u)}by part\n\nint{sin^-1(u)}\nlet m=sin^-1(u)                dv=du\n     =1/sqt(1-u^2)                u du\n=usinx^-1(u)-int{u/spt(1-u^2)}\nint{u/spt(1-u^2)} int this by subitution\n\nthen u will get\nsinxsin^-1(x)+(1/2)(1-sin^2x)\n\nso the final ans\nxlogsinx-sinxsin^-1(x)\n+(1/2)(1-sin^2x)+c\n\nplz try to answer this by yourself ,i hope i gave you the idea,thankx
601534
To convert from polar to cartesian co-ordinates, yousimply use:\n\nx = r cos (theta)\n\nand:\n\ny = r sin (theta)\n\nYou have radians - so make sure your calculator is set to radians.  This gives:\n\nx = 15 × cos (2.75) = -13.864...\ny = 15 × sin (2.75) = 5.7249...
677393
it is probably easiest to expand into a polynomial like bboy does. or use u substitution:\nlet u =1-3x\ndu=-3 dx\nso integral of -2/3 u^2 = -2/9u^3 = -2/9(1-3x)^3
840169
u= x\ndv=cosx dx\n\ndu =dx\nv= sinx\n\n\nintegral of xcosx dx = xsinx - integral of sinx dx\n= xsinx + cosx +C\n`
88038
It's been too long, so I'm very rusty on this but you could look at Lp norm  ||f|| = pth Root of (integral( f^p)) where the integral is taken over the real line. Try L2.
671750
Yes, click on window and put in the range that you want x and y to be.\nYou will have to push the green button to access the window function.\nNote: The domain applies to c and the range applies to y.
664950
In the polar coordinate system r is always positive. It´s the distance from the origin to the point. \n\nThe transformation to rectangular (or rather carthesian) coordinates is as follows:\n\nx = r cos (theta)\ny = r sin (theta)\n\nSo assuming (r,theta) = (5,52 deg) I find\n\nx = 5 cos (52 deg) = 3.08\ny = 5 sin (52 deg) = 3.94 (both rounded off to two decimals)
34052
I do not know a preprogrammed utility to do this, but can point you in the direction that you might need to go.  See the reference below for the math and an on-line converter:
463511
Use the double angle identities.