Please Help Me With This Calculus Exercise

[mufarrid]

Find the area enclosed by the curve r = 5cosθ

[JimF]

2π

[bobbinwales]

Is that right Jim?

[bobbinwales]

I was nearly top set maths 40 years ago and vaguely remember integrating and integral of cos x is sin x.. . But I don't know what the from and to range is... is it a full question.
How did you do yours Jim?

[JJ109]

How can the area be a linear amount?

[bobbinwales]

There's no units as such specified so answer don't need them either??

[bobbinwales]

Eg a circle of radius 1 (unit) has an area of just pi (units squared) I remember that much jj

[bobbinwales]

My guess is the answers just 10.... but it does depend where the curve start's and end's is it a full 180 degree range?
Is jomf around. Or pp or bienchico or Jim 360 ora mathamaticiam

[Etch]

Assuming that we are dealing with polar coordinates, the curve r = 5cosθ for 0 < θ < π traces out a circle of radius 5/2 centred at x=5/2, y=0. For π < θ < 2π the same circle is repeated. The area within the circle is therefore π.(5/2)² = 25π/4

[bobbinwales]

Best answer still marked as jimf
The word curve dosnt suggest a full circle to me but I can't argue with your etch 'cos' its over mead... a 'sine 'of old age

[Etch]

See here bobbinwales:
https://socratic.org/questions/59ee75b211ef6b09d0450a18#494508
In the OP's case the curve is 5 times larger, linearly, so 25 times larger area-wise. It seems the mods are too busy chatting about the BBC Quiz of the Year. ;-)

[bobbinwales]

Okay thanks etch will wait and see. The OP never mentioned polar coordinates but I never done those at school or at night school so am sure your right if thats what they meant.
I googled curve of r= 5cosθ and it gave me the expected wave graph curve I use to see in electronics rather than the circle you get from polar coordinates.

[jim360]

JimF's answer is wrong. The general formula for area enclosed by a polar curve in the form r = f(theta) is given at, eg, the following link:

https://en.wikibooks.org/wiki/Calculus/Polar_Integration

Here I'd choose a = -pi/2 and b = pi/2, (a = 0, b = pi will also do), and then you can integrate using tricks such as cos^2(x) = 1/2(1+ cos (2x)), or some such.

Etch's method is, however, by far the cleanest.

[melv16]

Scratches head big time over this thread:-)

[bobbinwales]

https://www.wolframalpha.com/widgets/view.jsp?id=d363321964240b3d58c776bddcab2be
this calculator gives a result thats double Etch's....maybe tehres an obvious explanantion

[Etch]

bobbinwales - as I mentioned earlier, θ only needs to go from 0 to π to create a closed circle from the formula r = 5cosθ, so to get the area enclosed by this curve the integration is performed just between those limits and not from 0 to 2π. Doing so using the WolframAlpha widget gives the correct answer - 25π/4

[bobbinwales]

Thanks etch... a stretch to far for me.
Can you help this one here
https://www.theanswerbank.co.uk/Science/Question1778675.html

[Peter Pedant]

thread in the manner of my maff master Fatty barrett
sort of restores my sanity