Video Transcript
The figure shows the steps to
producing a curve 𝐶. It starts as the boundary of the
unit square in figure (a). In figure (b), we remove a square
quarter of the area of the square in (a). In figure (c), we add a square
quarter of the area that we removed in (b). In figure (d), we remove a square
quarter of the area of the square we added in (c). If we continue to do this
indefinitely, we will get the curve 𝐶. We let 𝑅 be the region enclosed by
𝐶. By summing a suitable series, find
the area of the region 𝑅. Give your answer as a fraction. Is it (a) one-quarter, (b)
four-fifths, (c) one-half, (d) two-thirds, or (e) three-quarters.
The easiest way to deduce the
series that represents this process is to simply start with the area of the unit
square then add or subtract the next few terms. So with figure (a), we start with
the area 𝐴 equals the unit squares area, one. In figure (b), we subtract
one-quarter of the square, which has an area of one-quarter. In figure (c), we add back on
one-quarter of the area we just subtracted. So we’re adding one-quarter times
one-quarter. And finally, in figure (d), we
subtract one-quarter of the area we just added in (c). So we’re subtracting one-quarter
times one-quarter times one-quarter.
We can rewrite this as one minus
one-quarter plus one-quarter squared minus one-quarter cubed. And if we do this indefinitely, the
series will look like this, with each term being minus one-quarter times the
last. This is a geometric series with the
starting term little 𝑎 equal to one and a common ratio 𝑟 equal to negative
one-quarter.
Recall that the formula for the sum
of an infinite geometric series with a common ratio of magnitude less than one is
given by 𝑆 ∞ equals 𝑎 over one minus 𝑟. So again, little 𝑎 equals one and
𝑟 equals negative one-quarter. So this means that the area of the
region 𝑅 is equal to one over one minus negative one-quarter, which is equal to
four-fifths. Going back to our possible answers,
we can see that this matches with (b), four-fifths.
Moving on to the next part of the
question, consider the vector field 𝐅 of 𝑥, 𝑦 equals 𝑦, two 𝑥. What is the function 𝜕𝐅 two by
𝜕𝑥 minus 𝜕𝐅 one by 𝜕𝑦?
So we have 𝐅 one, the first
component of 𝐅, equal to 𝑦 and 𝐅 two, the second component of 𝐅, equal to two
𝑥. Therefore, 𝜕𝐅 one by 𝜕𝑦 equals
one and 𝜕𝐅 two 𝜕𝑥 equals two. Therefore, 𝜕𝐅 two 𝜕𝑥 minus 𝜕𝐅
one 𝜕𝑦 equals two minus one equals one.
Let’s keep some key information and
move on to the final part of the question.
Use Green’s theorem to evaluate the
line integral, the integral over the curve 𝐶 of 𝐅 dot d𝐫, where 𝐶 is the curve
above.
One last thing before we move on,
remember that we evaluated the area 𝐴 of the region 𝑅 to be four-fifths in part
one.
So at first glance, this doesn’t
seem to have much to do with the previous parts of the question. But consider 𝐅 dot d𝐫. This is the scalar product of the
vector function 𝐅 with components 𝐅 one and 𝐅 two and the line element d𝐫 with
components d𝑥 and d𝑦. Evaluating this, we get 𝐅 one d𝑥
plus 𝐅 two d𝑦. So the integral over the curve 𝐶
of 𝐅 dot d𝐫 is equal to the integral over 𝐶 of 𝐅 one d𝑥 plus 𝐅 two d𝑦.
Green’s theorem states that if we
have a closed curve 𝐶 enclosing a region 𝑅, then the line integral around 𝐶 of
two functions 𝐿 with respect to 𝑥 and 𝑀 with respect to 𝑦 is equal to the double
integral over the region 𝑅 of 𝜕𝑀 𝜕𝑥 minus 𝜕𝐿 𝜕𝑦 with respect to 𝑥 and
𝑦. Looking at our integral, we can see
that we have exactly this form with 𝐿 being 𝐅 one and 𝑀 being 𝐅 two. And we can indeed use Green’s
theorem as 𝐶 is a closed curve. So we have the integral around the
curve 𝐶 of 𝐅 dot d𝐫 is equal to the integral around 𝐶 of 𝐅 one d𝑥 plus 𝐅 two
d𝑦. And by Green’s theorem, this is
equal to the double integral over the region 𝑅 enclosed by 𝐶 of 𝜕𝐅 two 𝜕𝑥
minus 𝜕𝐅 one 𝜕𝑦 with respect to 𝑥 and 𝑦.
In part two, however, we already
found 𝜕𝐅 two 𝜕𝑥 minus 𝜕𝐅 one 𝜕𝑦 to be equal to one. So this is in fact just the double
integral over 𝑅 of d𝑥d𝑦, which is equal to the area of 𝑅, which we already found
in part one to be equal to four-fifths. Going back to our possible answers,
we can see that this matches with (a), four-fifths.
