• If you are citizen of an European Union member nation, you may not use this service unless you are at least 16 years old.

  • You already know Dokkio is an AI-powered assistant to organize & manage your digital files & messages. Very soon, Dokkio will support Outlook as well as One Drive. Check it out today!


Question 4

Page history last edited by Rence 15 years ago

A Calculator MAY be used to help answer this question.

Be certain to limit yourself to the four permissible calculator functions.


Let R be the region in the first quadrant bounded by the graph of y = 4 - x2/3, the x-axis and the y-axis.


(a) Find the area of region R.


(b) Find the volume of the solid generated when R is rotated around the x-axis.


(c) The vertical line x = k divides the region R into two regions so that when these regions are rotated about the x-axis, they generate solids with equal volumes. Find the value of k.


Quick Find Code For This Problem.


Legend: [QUES4LGND]

Solution For Part a.): [QUES4PA]

Solution For Part b.): [QUES4PB]

Solution For Part c.): [QUES4PC]


How To Use Quick Find.


-To Use Quick Find, copy (Ctrl+C/Command+C) the code (All the stuff INSIDE the square brackets in the code region above) for which ever section you'd like to go to, then, hit in this order;






or if your on a Mac You'd hit






Then, your webbrowser should bring you directly to the beginning of the solution indicated by the code, making navigation through longish solutions much easier. Enjoy :]



 Hi I am Benchmen and I will be solving this problem for everyone. I will be using different colours and font to solve the problem.




Blue - Steps taken


Red - Important to remember


Underlined and Italicized - Answers or other important numbers


Bold - Important terms


Purple - Other stuff



 a) This first part is simple. All we have to do is to find the area under the curve. To do that we will have to integrate the given function. So to integrate we need the limits of integration. According to the problem, the area is bounded by the y-axis. That means one of the limits is x=0. We need to find the other limit. Since the curve is bounded by the x-axis, the other limit must be a zero/root because that is where the area "closes up".



Let's find the root.


Let's look at the function on the Cartesian Plain. You don't really need to look at the graph at this point because the graphing calculator has it's limits in terms of resolution. But it may be useful to understand the behaviour of the graph. This will be our graphical approach to finding the root of the function.



Well it looks like there is a root at x = 8, but we don't know for sure so let's confirm by finding it algebraically.





To find the roots algebraically, make the function equal to zero.




Make the function equal to zero to find out what the zero is.







Okay now we know that the function has a root at x = 8. That is our second limit of integration.


Now let's find the area under the curve. So all we have to do is integrate the function from 0 to 8.





So what I did here was I antidifferentiated the given function. To anti-differentiate the function, reverse the power rule for each term.

The rule to reverse the chain is:

1.  Add 1 (one) to the exponent

2.  Divide the term by the new exponent


Remember that the Second Part of the Fundamental Theorem of Calculus says that a function integrated from a to b is equal to the change of the parent function (antiderivative) within the interval a to b. I'll make an image explaining this in algebraic terms.


If f(x) = F'(x)



Now you know why I antidifferentiated the given function. Now let's continue with applying the Fundamental Theorem of Calculus. So just substitute the limits to it's corresponding variable and you will get the answer.




Notice that if you are solving a polynomial for x=0, it is most likely going to have an answer of 0.




Therefore the Region has an area of 64/5 units2 or 12.8 units2


Now we have the answer to part a). Don't forget the units. Let's do part b)



 b) This part involves finding volumes of revolving areas. Here is a little video to review Solids of Revolution.




Okay now that you know what a solid of revolution is, let's solve this problem. The first step would be to visualize a sample section/slice, in this case, a sample disk. The disk would be a very thin cylinder. Let's look at the formula for the volume of a cylinder.



The sample that we took has a volume equal to this formula. Let's use this formula to make a general equation that represents the volume of the whole solid. Let's see if we can make this formula into a function of one variable. If we look at our sample slice, the radius of the sample solid is the distance from the x-axis to the curve. That means the radius is the height of the function, or the value of the function. Now for the height. Remember we are going to add a lot of thin disks, an infinite amount of disks to be exact, so they must be small, so small that the changes in x will now be called dx. Here is an image:




Since we will be adding an infinite amount of these thin disks, we will integrate that function between the interval. The integral sign is a modified "S" that represents an infinite sum of things within an interval. So let's add an infinite number of these disks between 0 and 8.





Notice that I brought the pi outside the integral because it is a constant multiple. You don't have to include a constant when integrating.



In this step, I expanded the squared binomial so I can produce a polynomial because polynomials are easier to antidifferentiate.



Similar to part a), just antidifferentiate the integrand and substitute according to the First Part of the Fundamental Theorem of Calculus.



Remember since the x will be substituted for 0 and the function is a polynomial, the result will be 0 so just worry about the 8. Just evaluate and you should get:



Therefore the volume of the solid is 1024π/35units3 or approximately 91.9140 units3.  Don't forget the units.



c) This question is asking for the upper limit that the volume function should be integrated that will produce half the volume from the last question. So let's find out what half of it is.



Therefore half of the volume from part b) is 512π/35 units3 or approximately 45.9570 units3. Let's go back to the integral we made earlier for finding the volume and use it to find the upper limit.




So this time we are looking for the upper limit so we will use a variable for that. I'm going to use the letter k. You can use anything you want, you could even use GEORGE but that would take too long to write down during an exam =P. Next just integrate normally.







Don't worry about the 0 here again. Here the pi's reduce.





Once you arrive at this point, you can use your handy-dandy calculator function to find the zeros/roots of that equation above. Insert the above equation into Y1.


Since this is an exam prep, I guess I should include that there are 4 graphing calculator functions that are permissible in the exam other than the regular calculator functions (+, -, x, ÷) and storing large, small, or non-terminating numbers :


1. Graphing functions (Y=)

2. Solving equations or finding zeros/roots (2ndF: Calc: Zero)

3. Numerically integrating a function (2ndF: Calc: ∫f(x)dx)

4. Numerically differentiating a function (2ndF: Calc: dy/dx)



Here is an image of the function.



Then go to 2ndF: Calc: Zero. You should get:





Therefore to get half of the volume from part b), you have to integrate the volume formula from 0 to k which is approximately 1.3605.




Now you have just solved Questions 4.


Thank You Sitmo Equation Editor, Fooplot.






Comments (0)

You don't have permission to comment on this page.