Find area inside this loop!?

Question

Ok the curve, x = t^2 and y = -t ln (t^2) has a loop that begins and ends at x intercepts. Find the area inside this loop.

Answer 1

Put y = 0

this gives t = 1

Thus the point is (1,0)

Now integrate.

Answer 2

All of your answers are correct. You could solve this from θ=-π ⁄ 2 to π ⁄ 2, or realize the symmetry and double the value you get going from θ=0 to π ⁄ 2. But... There is a missing piece that I'd like to make certain you understand. Their integrals are all set up correctly and there is some reasoning for that. You asked for "any insight." It would have helped a lot if you'd shown some of your work so that others could see why you were getting negative answers. You get even better help, then. However, I'll take a different path here and see if I hit the mark. How is it that they came up with some ½ r² dθ? Was it arbitrary (just out of the thin air???) No. Imagine your integration taking place. There is a line that is 'r' long, positioned at an angle of θ. Now, you need to allow the line to sweep out a very, very small (infinitesimal, really) area. To do that, the angle is moved just a very tiny bit, moving from θ=θ₀ to θ=θ₀+dθ. So draw out two lines on your sheet of paper so that you can see both of them with a tiny angle between them and label the narrow angle, dθ. Now ask yourself... how do you figure out the area swept there? Well, since the angle dθ is infinitely small, you can consider the triangle formed there to be a right triangle, with one of the legs being 'r'. But the very tiny perpendicular is how big? Well, it is r⋅ sin(dθ). But since dθ is so small, it is the case that sin(dθ)=dθ. So it's length is just r⋅dθ. So the area of the infinitesimal triangle is ½⋅(r)⋅(r⋅dθ), or just ½⋅r²⋅dθ. And you integrate up those tiny slices, as you walk θ from some value to another, dθ at a time. Just in case that part hadn't been made clear enough to you. Hope it helps.