It looks like you want to compute the double integral
over the region <em>D</em> with the unit circle <em>x</em> ² + <em>y</em> ² = 1 as its boundary.
Convert to polar coordinates, in which <em>D</em> is given by the set
<em>D</em> = {(<em>r</em>, <em>θ</em>) : 0 ≤ <em>r</em> ≤ 1 and 0 ≤ <em>θ</em> ≤ 2<em>π</em>}
and
<em>x</em> = <em>r</em> cos(<em>θ</em>)
<em>y</em> = <em>r</em> sin(<em>θ</em>)
d<em>x</em> d<em>y</em> = <em>r</em> d<em>r</em> d<em>θ</em>
Then the integral is
Answer: The answers are
(i) The slope of segments DE and AC is not 0.
(ii) The coordinates of D and E were found using the Midpoint Formula.
Step-by-step explanation: We can easily see in the proof that the co-ordinates of D and E were found using the mid-point formula, not distance between two points formula. So, this is the first flaw in the Gina's proof.
Also, we see that the slope of line DE and AC, both are same, not equal to 0 but is equal to
which is 0 only if
So, this is the second mistake.
Thus, the statements that corrects the flaw in Gina's proof are
(i) The slope of segments DE and AC is not 0.
(ii) The coordinates of D and E were found using the Midpoint Formula.