Man thank you so much, literally the first time i actual properly understood polar coordinate transformations despite trying to wrap my head around it for like 3 months. Appreciate you!!!
@marcuswauson The integrand is [cos(theta)-(1/2)*cos(theta)] but when integrated, it becomes [sin(theta)-(1/2)*sin(theta)] from zero to pi/4. When evaluated equals sqrt(2)/2 - sqrt2)/4 = sqrt(2)/4
In my school they simply say: evaluate the following integral just as given in part (a) above. It is entirely ALL UP TO YOU how you approach it. Of course one method makes life a whole lot easier than say compared to another one. It is expected that you know which method to use accordingly.
makes me realise how much I have forgotten in the 40+ years since uni. Expressing the integral in that graphic form was not something we were taught and it really helps me to grasp the whole concept; would have grokked so much more if we had been!
Thanks a lot, I was struggling to solve a problem related usage of polar coordinates in my assignment, but now i solved it in just 5min. thanks once again!
yes; and this is how we deal with it. find the range of angles for which r is negative, split these segments away from your first integral and put a negative sign in front in order to find the area. example r=cos(theta) from 0 to 2pi gives negative r from pi/2 to 3pi/2. the area enclose by r=cos(theta) = integral of rcos(thta) from 3pi/2 to pi/2 plus (-) integral cos(thta) from pi/2 to 3pi/2. integrating the original function from o to 2pi gives 0 if you ignore the fact r goes negative.
@PeaceUdo Question a and b) The upper bound for y is y=x. The line y = x is always at a 45 degree (pi/4) angle with the x axis. If you dont get why, then for example lets say y = x = n (as y=x) then tan θ = n/n tan θ = 1 therefore θ=45 degree (pi/4)
Because he has (1/r^3) r drdΘ, if you combine that r with (1/r^3) it would have a negtive power if you put it over one, (r has a power of one), therefore it becomes (1/r^(3-1)... and becomes 1/r^2. Hope this helps.
Not only do the brilliant students of MIT get excellent student teachers, but I can't understand majority of what my teacher says through his thick accent...and no my school is not well known even in the city that it resides in. Thanks David and MIT!
only in Calc BC trying to solve an argument with a friend when i found this… video made the concepts simple to understand and was extremely informative!
cos of 0 is not 0, cos of 0 is 1, your answer to a should be (squareroot of 2) minus 2 all over 4. As cos of 0 is one, and one minus a half should give you squareroot of 2 over four minus a half.
Students note double or triple integration does not give correct results. Follow simple integration. The question is why count something twice as it happens in double triple integration.
the one thing i think that is missing when converting is that the F function at the end is not a function of x, y but r, theta you cant just plug in f given a function of x,y when in polar form
Thanks! I think you articulate it really well. That's awsome! Wish you're my prof. Not to be disrespectful, but are you an undergraduate? You look like a peer.
to prove this , we use jacobian method. And we all know that from cartessian to polar coordinate , x=cos@ and y=sin@ then dxdy change to drd@ and apply this to jacobian matrix (dx/dr dx/d@ , dy/dr dy/d@) then we get rdrd@
It was subtle but you have to watch out for the 1/r^3 r dr dtheta. The r in the nominator position crosses out one of the rs in the denominator leaving 1/r^2 dr dtheta.
Does the last example diverge? Because the the internal integration integral[r=0,r=2sin(theta)] r^-2 dr = -1/r | [r=0,r=2sin(theta)] = -(1/2sin(theta)-1/0) = diverge
I think there is a little mistake in the end of letter a). The result should be sqrt (2)/ 4 - 1/2, instead of sqrt (2)/4. Thank you for the excelent explanation.
No. There are several reasons why. First, r is a length, which logically can't be negative. Second, If you look at the formula, r=(x^2+y^2)^(1/2), you'll realize that there's no way for it to be negative since it's components are all squared.
