Solve This plox my Exam is going on

Answered question

2022-01-29

Solve This plox my Exam is going on soo 

Answer & Explanation

user_27qwe

user_27qwe

Skilled2023-04-22Added 375 answers

To evaluate the integral ∫Cz2dz, where C is the curve y=3-x2, we first need to parameterize the curve.

Let's set x=t, then y=3-t2. The curve can be written as C:(t,3-t2), where -3≤t≤3.

Now, we need to express dz in terms of dt. We have z=x+iy=t+i(3-t2). Taking the differential of z with respect to t, we get dz=dx+idy. Substituting x=t and y=3-t2, we have:

dz=dx+idy=dt+i(-2tdt)=(1-2it)dt

Now we can evaluate the integral using the parameterization and the differential we just found:

∫Cz2dz=∫-33(t+i(3-t2))2(1-2it)dt
             =∫-33(t2-6it3-8t4+9it-9it5)dt

We can now integrate term by term, using the fact that ∫idt=it+C:

∫Cz2dz=[t33-3it42-8t55+9it22-9it66]-33

After plugging in the limits of integration and simplifying, we get:

∫Cz2dz=72i5

Therefore, the value of the integral ∫Cz2dz over the curve C is 72i5.

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