Calculus: Area problem?

Question

Find the area in the first quadrant bounded by the arc of the circle described by the polar equation r=2sin(theta)+4cos(theta).
okay i would use the formula dA=r^2 d(theta). Thats what i used is this correct? and when i use this i get: 4sin^2(theta)+16cos(theta)sin(theta)+16cos^2 (theta). The next step would be to integrate right? But when I Integrate this... is there a way i can put the 4 from 4sin^2 (theta) to the left of the integration sign. like in an example r^2/2 the 1/2 is put to the left of the integration symbol and r^2 is integrated (is this making sense to whoever is reading this). well any way is there a way to put the whole numbers like the 4,16,16 to the left of the integration? if not maybe someone can help me do it there way? please i really need this question. please show some steps. i just really want to understand how to be able to integrate!! thanks i know that was long!! i really appreciate everyone's response!!!!=D

thanks everybody for responses. Dr.spock ur explaination was very good. i had 5pi/2+4 but when i recieved a post bout it being 5pi+8 i second guessed myself. so we all agree its 5pi/2 +4 ?

Answer 1

Mea culpa! I'm taking back a statement that the shape is not a circle. It is indeed a circle, with equation:

(x - 2)^2 + (y - 1)^2 = 5,

that is a circle with radius sqrt(5), centred on the point (2, 1). It intersects the axes asymmetrically, at (4, 0) and (0, 2).

However, doing the double integral with area element dx dy is a bit tricky because the limits are awkward. It's much better to go with the polar coordinates; that is just a single integral.

[The fact is (as rhsaunders pointed out), the question of HOW to work out the area is not only simplest geometrically, it is in fact TRIVIAL. I'll do it below,*** after completing the calculus treatment. But it wasn't necessarily clear whether your question, "Calculus: Area problem?" meant "Here is an area problem. SHOULD I do it by calculus?" If that was your original question, as posed to you, the answer should be "No, do it geometrically!" However, you seem to have some calculus issues/questions, so I'll continue with that approach.]

One minor but nevertheless significant problem (a writing slip?) is this: dA is HALF of what you wrote, that is (1/2) r^2 d(theta). Why?: The area of an elementary triangle is "1/2 base x height" --- the "base" is r d(theta), the "height" 'r'. (That's why, when integrating over a full circle, you get (1/2) x r^2 x 2 pi = pi r^2 for the area of the circle.)

Let's call "theta" 't', for goodness sake! So the integral is:

1/2 Int (0 to pi/2) of [4 sin^2 t + 16 sin t cos t + 16 cos^2 t] dt.

(Note: Since you asked, multiplicative constants can always come outside the integral sign. Personally, I tend not to do it explicitly; I rather concentrate on integrating the variable part, and then mentally multiply by the constant.)

A useful fact is that the "average value" of sin^2 t or cos^2 t betweeen 0 and any multiple of pi/2 is exactly 1/2. So the square terms inside the brackets simply give us [(pi/2)(4 + 16)/2] = [5 pi]. The remaining term is the integral of [8 sin (2t)] = [ - 4 cos (2t)] evaluated between t = 0 and pi/2, or + 8 if I haven't made a mistake; alternatively. it's [8 sin^2 t] evaluated between t = 0 and pi/2, also 8, thank goodness. So the integrated [...] brackets give us [5 pi + 8], and therefore the area is half that, or:

2.5 pi + 4 or ~11.85398... .

What did you get?

Live long and prosper.

P.S. 'gianlino', like you, seems to have neglected the (1/2) factor in dA. Otherwise, we would have agreed. (His result is exactly twice mine because of not using that (1/2) factor.)

*** P^2.S. As stated above, it's geometrically trivial. The two axis intersections and the centre of the circle are all on a diameter. Therefore the area is the sum of a tilted semicircle [radius sqrt(5)] above that diameter plus the area of a right-angled triangle with perpendicular legs 2 and 4. Hence it's:

(1/2) pi r^2 + (1/2) 2 x 4 = 2.5 pi + 4 (!)

So rhsaunders was quite right. The calculus approach is like using a sledgehammer to crack this particular nut, once one has appreciated what the shape and disposition of the curve are. But at least this confirms the correctness of the calculus result.

Answer 2

For the first quadrant int of sin^2= int of cos^ 2 = pi/4 since the sum of the 2 is pi/2. For the cross product term, int of sin cos is sin^2 /2, so that gives 1/2. I guess you can add up eveything and get

4 pi/4 + 16 /2 + 16 pi/4 = 5 pi + 8

Answer 3

D. its spectacular! right here is why: S(-3x^2+12)dx now combine to get -x^3+12x | -2 to 2 so: [-8+24]-[8-24] which equals 32. A is mindless. in case you combine a spinoff, you get th comparable function returned (-3x^2+12). B is incorrect because of the fact the consistent of integration is cancelled out in a different critical C. is mindless. you do not ought to element out a destructive, the mathematics tells you the area.

Answer 4

Consider converting to Cartesian coordinates. That should make things considerably more obvious, and you can probably solve the whole thing geometrically.