parametric question

How do I know whether I should use t is -3 or t is 0 for the lower bound of the integral? I thought to use t as -3 as it seemed to be the lowest point where the curve crosses the axis on the graph but this gave me the area above and below the ellipse? Should it not just give me half the area above the ellipse?

When t is negative large, y is positve large (quadrant 1), so as t increases from -inf it starts in the top right, passes through B at t=-3, then A at t=0 then B again at t=3 and heads off in quadrant 4 as t->inf. So t going from 0 to 3 should give the positive part above the x axis then double it.

Use t =-3 as the upper bound

When you are writing your limits in terms of x, you get -4 and 41
So the 41 is upper limit and -4 is lower

Changing your limits to t, means 41 becomes -3, so that is upper even though it’s negative.

Btw what did the markscheme say R was?

thank you

thanks I get it now. I think you use t = 3 as upper tho. it said 648