please calculate me number 5 and number 1, for number 5- there is some partial calculations, please continue calculating for me until it gets the final answer. For question number 1- the answer is 11572.5 pi ( please show me the calculation for that), Thanks.

5. A unit disk centered at the origin is sliced so that the right portion has width h. Derive a formula A(h) for the area of this slicer To find appropriate antiderivatives use WolframAlpha. Show verication that A(O) = n and A(2) = 1r. Solution. Let us place the unit disk in the coordinate system. The boundary of the unit disk is given by the equation of the unit circle as\" + y\" = 1, so the positive part is the graph of f(:c) = W. The shaded smaller slice is then twice the area under the graph of f(x) = m between r : 1 h and 1: 1 A=2/ Vlmgdm. 17). So far we have not studied techniques (other than guessing) to solve the above integral. We can check the answer with Solution. Let us place the unit disk in the coordinate system. The boundary of the unit disk is given by the equation of the unit circle z2 + y2 = 1, so the positive part is the graph of f(z) = m. The shaded smaller slice is then twice the area under the graph of f(z) = between a: = 1 7 h and 1: 1 A:2/ Vliaridz. 17}, So far we have not studied techniques (other than guessing) to solve the above integral. We can check the answer with Wol'amalpha that /V1z'2da:= % (Ivlzz-larcsinz) +C. Finally, A = 2 [% (aim + arcsin 12)]: 7h 7 g 7 (1 7 rot/1 7 (17 h)? 7 arcsin(1 7 h). One way to rest assured that the formula is correct is to check it when h = 0, that is when we get the right half disk. We should get the area of one half of a unit disk, which is 1r/2i 1. Find the volume of the solid obtained by rotating the region bounded by the curves y : I2 and y = and the lines a: = 4 and .1- : 9 about the aaxis