Question
4_ Consider the integral$ fce)a:over a closed, positively oriented contour of the function f (2) = 24 + 2. (a) By computing the antiderivative of f (2) , that is the function F(2) such that F' (2) = f (2), show that the integral is zero [3](b) Let the closed contour be the unit circle Izl = 1_ By employing a suitable parametrization of the circle, ~(t) , compute the integral once more and show that it is zero: [5](c) Using the same function f (~) as defined above, find all solutions to the
4_ Consider the integral $ fce)a: over a closed, positively oriented contour of the function f (2) = 24 + 2. (a) By computing the antiderivative of f (2) , that is the function F(2) such that F' (2) = f (2), show that the integral is zero [3] (b) Let the closed contour be the unit circle Izl = 1_ By employing a suitable parametrization of the circle, ~(t) , compute the integral once more and show that it is zero: [5] (c) Using the same function f (~) as defined above, find all solutions to the equation f(2) = 2+26)34+ (-i)44 and write them in polar form_ [6] (d) Use the residue theorem to compute the integral dz 24 + 1 on a positively oriented square with vertices {-2, ~2-2i ,0, -2i} . Sketch the square and the poles of the integrand in the same graph. Simplify your result as much as possible_ [6]


Answers
a. Let $B$ be the region between the upper concentric hemispheres of radii $a$ and $b$ centered at the origin and situated in the first octant, where $0 < a < b$.spherical coordinates $\quad(\rho, \theta, \varphi) \qquad$ is $F(x, y, z)=f(\rho) \cos \varphi . \quad$ Show that $\quad$ if
$$
g(a)=g(b)=0 \text { and } \int_{a}^{b} h(\rho) d \rho=0,
$$
$$
\iiint_{B} F(x, y, z) d V=\frac{\pi^{2}}{4}[a h(a)-b h(b)]
$$
where $g$ is an antiderivative of $f$ and $h$ is an antiderivative of $g .$
b. Use the previous result to show that
$$
\iiint_{B} \frac{z \cos \sqrt{x^{2}+y^{2}+z^{2}}}{\sqrt{x^{2}+y^{2}+z^{2}}} d V=\frac{3 \pi^{2}}{2}, \text { where } B \text { is }
$$
the region between the upper concentric hemispheres of radii $\pi$ and 2$\pi$ centered at the origin and situated in the first octant.
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So having this problem we evaluate a double integral by changing to polar cornets. And the given integration is the double and triple over the region. R. Y squared over X squared plus y squared D. A. Where the region R. Is a region that lies between the circle X squared plus y square is equal to a square and X squared plus Y square is equal to be square. Her. A. Is less than B and A. Is greater than zero. So this is the condition we are given with. Now first I'll be sketching this area. So the shared region, this one by these lines are the is a region over which we have to evaluate the interval. The outer circle represents the equation X squared plus y square. As it calls to be square and the inner circle it rebellions. This is determinate here. This is all and inner circle represents X squared plus y square is equal to a square, notice all this. Uh Androgel, we had to change into polar cornett. And for polar corn it's you know, they're X. Is the calls to our cousin Tito why it equals to our science theater and D. A. Began right this thing us are D R D Tito. How can you use this concept here? So the double integral would become equals to the double integration. Why square would mean our square science where heroes are square, sine squared, Tito, divide by X squared plus y squared, which is equal to our square X squared plus y squared would be in our square dot. D A V F R D R D T. To if you have any confusion that how we are able to write X squared plus Y squared as our square. You can put X as here are strengthened and Y. S. R. Sorry, excess arcosanti, A and Y S. R. Santa. You'll get our square. No, this Oscar and Oscar will cancel this one. And this one next we have to put the limits for R. And theta. And here we see that the radius R. Is wearing between a baby and A. So we can say that art is ranging between A. And B. So the limit would be A to B. The 4th ERA. It is 0 to buy here. The zero and again in the coming here zero to cuba zero to buy. Now we can easily integrate this. So it will be equals two LTD 0 to 2 pi. No Science square to we are integrating with district are so Science Square theater would be concerned and our dear. So it would become our square divide by two and the liberties from A to B. And the hero. Now we can put the limits so it will be called to 0- two Pi. Science quoted as it is And this would become be square minus a square divide by two duty to. And this time we can take outside the integration. So it would become be square minus a square divide by two. Integration. 0 to 2 pi By using programmatic formula we can write that Science where Tito it is equally true. 1- Co sign. Don't try to divide by two. So this would become one minus cuisine duty to divide. But we just bring Science closer to in this form not dated them and we can take you outside this integration. So it would become for here. No again easily integrate this So it will be equals two. B squared minus a square divide by four. And here it would become tita minus the integration. Of course I intend to is signed to teeter divide where to So sign to teeter or divide by two And the limit is from 0 to 2 pi. No, simply we have to put the limits so be square minus a squared divided by four A Parliament. We have to buy minus sign two times two pi would become forward by divide by two minus the lower limits alone. But we have heard is zero And signed 0- 0- zero as it is. Well for to simplify this and get our results to be pi divided by two, be square minus a square. So this is a required answer. And finally we can conclude that the double integration oh y square over X square Bless Y square D A over the region R. Is the calls to. Why did World War Two be square minus a square? So this require answer that was asking the question. I hope you have understood the problem. Thank you.