Question

http://prntscr.com/a2tmfl

Answer 1

@freckles

Answer 2

ok so we want to find the area bounded by f and g and y=1

Answer 3

so let's see if there is any intersection between f and g first

Answer 4

\[f=g \\ \frac{y}{\sqrt{16-y^2}}=0 \\ \text{ so we have one intersection } y=?\]

Answer 5

0

Answer 6

cool

Answer 7

u=0 aka g(y)=0
anyways
you can draw u=y/sqrt(16-y^2)
make a rough graph of this
but this is all above the y-axis on the interval we care above
so we are kinda ready to compute our integral

Answer 8

or setup our integral at least

Answer 9

the lower limit is y=?
the upper limit is y=?

Answer 10

lower = 0
upper = 1

Answer 11

cool so the lower limit is y=0
and the upper limit is y=1

so you integral is
\[\int\limits_0^1 \frac{y}{\sqrt{16-y^2}} dy \]