(i) Yes. Simplify .
Now compute the limit by converting to polar coordinates.
This tells us
so we can define to make the function continuous at the origin.
Alternatively, we have
and
Now,
so by the squeeze theorem,
and approaches 1 as we approach the origin.
(ii) No. Expand the fraction.
and
are undefined, so there is no way to make
continuous at (0, 0).
(iii) No. Similarly,
is undefined when .
Answer:
∫ C ( y + e√x) dx + ( 2x + cosy² ) dy = 1/3
Step-by-step explanation: See Annex
Green Theorem establishes:
∫C ( Mdx + Ndy ) = ∫∫R ( δN/dx - δM/dy ) dA
Then
∫ C ( y + e√x) dx + ( 2x + cosy² ) dy
Here
M = 2x + cosy² δM/dy = 1
N = y + e√x δN/dx = 2
δN/dx - δM/dy = 2 - 1 = 1
∫∫(R) dxdy ∫∫ dxdy
Now integration limits ( see Annex)
dy is from x = y² then y = √x to y = x² and for dx
dx is from 0 to 1 then
∫ dy = y | √x ; x² ∫dy = x² - √x
And
∫₀¹ ( x² - √x ) dx = x³/3 - 2/3 √x |₀¹ = 1/3 - 0
∫ C ( y + e√x) dx + ( 2x + cosy² ) dy = 1/3
