@mmcdara 

No nothing mentioned about this. I dont know why, but they really are a bit vague in this book series. Any way the math you mentioned is the sort of math you get before you go to a school where you will learn this math book series. 

Any way, they sort of leave you in the dark and you have to figure out that all of a sudden you need that sort of skill to make it to that result mentioned. But they never did that sort of manipulation in the examples. So its a bit odd. 

Im having a bit of trouble seeing when to think outside of the box with math. Remembering the teachers back then, they to were sometimes a bit astranged by the answers the book gave. Its like lightning on a sunny day with blue skies... 

Im still not seeing how you got to where you came to. You can call it what you want, it just does not "show" to me that how it is done. Now i did a lot of books, and when they are good books, you are not left with questions. These books are really so so... 

At one school the teacher told us that they did not have this "part 2" from which this question came because then a lot of students would just walk off because the book was that bad, and they had their own printed book by the school that was much better written and easier to understand. So that really says something about this book. Im just using it to get good at maple. 

Any way thanks for the answers.

Greetings,

The Function 

@dharr 

Well i already tried this. It gave me the same results as you gave, and it did not look anything like the results shown by the book. So i was confused about it, and that is why i came here. But since you got the same results. What is good?

Greetings,

The Function

@The function 

I really dont know what a "pochhammer" is. Ive certainly havent seen any (yet). Is at a hammer to hit an anvil with? 

Greetings,

The Function

Well this is what i came up with:

But the answers are not pretty at all (in my opinion). How do it wright down "pochhammer"? I dont think they will count that as a pass on any exam. 

The book gave this as an answer: 

The formulas given in this paragraph:

The x=0 formula:

The x=a formula:

I really dont know what to think. Should ive looked further before asking, well maybe, but the answer i came up with still is not the one that is good enough to get to a point where you could say that its a good result. It is a right answer. But the answer i got from @Carl Love looks way better. 

Greetings,

The Function

Download Mapleprimes_Book_2_Question_1.1.mw

@Carl Love 

Hello Carl!

Well it seems to give the right answers. I will have to look into how it does this. The "subs", and "FormalPowerSeries" commands i did not know yet. I will use this while it does seem to work. But it feels kind of being on a raft on the ocean... Untill i understand what is going down. Handy. 

I hope that one day i will be as good as you are with maple. Maple seems to do it all, but without some guidance it is as good as a brick. 

Thank you!

Greetings,

The Function. 

@Carl Love 

It seems that you have found the final solution to this problem. It really does look good. 

I will study this some more tomorrow. 

Thank you!

Greetings,

the Function 

@Rouben Rostamian  

I already told you what it is. That is the solution. The shape of the sphere is already described by the formulas and the smartplot. We know the deal rho*g*the integral(going from the start of that part to the end of that part) of the formula for the ratio(ranging from 0 to 1) times the total volume of that part. And that is the solution. Right? That is my solution, i just want know what other people might come up with. 

You did not give me any good calculation on the solution of part .b and .c. The Pi is not what i saw earlier in your comment. So what is the deal? How did you come to those solutions?  

Greetings,

the Function

@Carl Love 

Ive thought of a lot of things, also taking it as an infinite number of circles that go up in size, everything. The thing is, you just cant get a good formula that does it. So i took the total volume and used the value of the formulas of .b and .c to know how much of the value should be counted. Its the one and only solution that really works, that I could think of. 

If you got something, proof it too me instead of a lot of good sounding words. Words are no solution to this matter. 

Greetings,

the Function

@Rouben Rostamian  
For a.:
That is the explanation. For ever height down, the radius grows with 1/4 the height. The formula for the volume of a cone according to google is: V=Pi*(r^2)*h/3, and that gives the formula volume V=Pi*((x/4)^2)*x/3. Its as easy as that. Nothing more to explain there. Im not going to wright a scripture about one question. 

For b.:
For the sphere it took the formula i got from a question i asked earlier. solve(x^2 + y^2 = 4, y), that gave sqrt(-x^2 + 4) and -sqrt(-x^2 + 4). That was for a circle with a radius of 2. I rewrote the formula to fit a circle radius of 1, and a rewrote it so that it fitted to what i needed on the x and y axis. Then i took the volume of a sphere: V= 4/3*Pi*r^3. Then I took 1/2 the sphere for the first part, and then the other half for the second part. Then i multieplied it with 9.81 with both parts. Multiplied the integration of the parts of the formulas relevant to the solution with 1000 and then got the answer in Joule. 

For c.:
The earlier part of the bottom of the sphere was rewritten to fit the situation this time. It was used as a ratio for just one half a sphere with a redius of 2 meters. The half volume of a Sphere was used, because half a sphere was asked for, and then i did the same thing as i did with b. 

Also i am not going to do some promotion scripture on how to do this, i think the explanation here is good enough. 

It took a while to figure it out. But in the end it was not too hard at all. You just need a lot of repetition. 

That actually reminds me of a guy called Wernher von Braun, he started as an amateur rocket enthusiast, and he was really bad at math. But when he realized math was what he needed to achieve his goals, he started to do a lot of math. In the end he became one of the greatest rockets scientists of all time!

Greetings,

the Function. 

@Rouben Rostamian  

For the formula of the first answer a. I used the ratio that is 2/0.5=4, so for ever part x it goes down, the radius increases with x/4, and that gives the formula. The shape does not change only the size. 

Your syntax looks like rocket science. :P Can i get your Maple sheet? So i can see what you programmed for the rest of answers too. :) 

I actually expected more people to answer. But this was a particularly hard one to answer. 

The book gave these answers: a. 0.82*Pi kJ (2,576.10J), b. 13*Pi kJ (40,840.70J), c. 39*Pi kJ (122,522.11J). 

SOMEHOW MY ANSWERS ARE NOT BEING DISPLAYED. IM VERY DISAPPOINTED DUE TO THIS. CAN SOMEBODY FIX THIS? ITS VERY HARD TO GET MY POINT ACROSS WITHOUT THE MAPLE SHEET BEING DISPLAYED.

Download Question_for_maple_primes_8.1.mw

For b. i got 41092.03192J, for c. ive got 129094.4256J. Its not too far off, but its not spot on either. So that is why i wanted to shed some more light on this matter. I still hope there is someone around that can show me what is up. 

@Rouben Rostamian  

I should try to do math earlier in the day. Now its the end of the day. Yesterday i was sleep drunk, just could not see anything that would look like an answer. 

Seeing your answer now, man, i should really try harder, the solution you gave actually is pretty obvious. How could i not see this?? :P 

Getting just the y on one side and the rest on the other side gives something that looks like a "normal" function. It is actually luck that it has a root, otherwise the half circle may not even have been possible. (Would they otherwise ask this question? Probably not...) 

Any way.

Thank you!

Greetings,

The Function 

@Kitonum 

I really dont know why your function plotting worked the way it did, i would expect it to go through the x-axis at Pi/2 and Pi, but it however crossed it at x=2. So odd. 

So it actually creates the cosine and sine circle. And you multiplied it with 2, thus giving the half circle. 

Really interesting. See, if you dont know what good input is. You could try all day in vain. Really helpful answer. Thank you!

Greetings,

The Function. 

@Ronan

That is still having the bottom of the circle, but it does not shot is how it is drawn. 

So that is not what i am looking for. Thanks for the answer though!

Greetings,

The Function  

@The function 

Going through it again, using factor(%) works works with some of the questions its solutions, but definitely not all of them. Not with the one i showed. 

Greetings,

The Function 

@acer 

You answer looks something similar to what my mathbook states the following (its in Dutch but if you only read the code you will get what is happening: 
 

But i want to know if there is any way to simplify the really long answer because that is what i am looking for. Not working around the long answer by making a more smart way to do the calculations.