2448 Reputation

11 Badges

14 years, 196 days

Dr. Robert J. Lopez, Emeritus Professor of Mathematics at the Rose-Hulman Institute of Technology in Terre Haute, Indiana, USA, is an award winning educator in mathematics and is the author of several books including Advanced Engineering Mathematics (Addison-Wesley 2001). For over two decades, Dr. Lopez has also been a visionary figure in the introduction of Maplesoft technology into undergraduate education. Dr. Lopez earned his Ph.D. in mathematics from Purdue University, his MS from the University of Missouri - Rolla, and his BA from Marist College. He has held academic appointments at Rose-Hulman (1985-2003), Memorial University of Newfoundland (1973-1985), and the University of Nebraska - Lincoln (1970-1973). His publication and research history includes manuscripts and papers in a variety of pure and applied mathematics topics. He has received numerous awards for outstanding scholarship and teaching.

MaplePrimes Activity

These are answers submitted by rlopez

Carl Love has provided a complete solution based on first principles. There are three built-in tools that would have drawn the region of integration and provided an appropriate integral for the volume inside the sphere but outside the cylinder.

The VolumeOfRevolution tutor in Student Calculus1 will implement the method of shells for any vertical axis of rotation.

There are two relevant Task Templates in Tools/Tasks/Browse: Calculus-Multivariate/Integration/Visualizing Regions of Integration/3D, one for integration in cylindrical coordinates, and one for integration in spherical coordinates.

See the attached worksheet for details.Tools_for_Volume.mw

The CenterOfMass command in the Student MultivariateCalculus package will provide the CM for both 2D and 3D regions that can be swept by a single mulltiple integral. Give the density function and the ranges of integration as equations of the form x=a..b, y=c..d, etc, and Maple returns either the unevaluated integrals, of the values for the coordinates of the CM.

The dchange command in the PDEtools package will also do it.


q:= (x^(5/3)-5*x^(2/3)) ;

Q:=convert(q, surd);


Roots are principal roots, and can be complex numbers. To obtain the real root in these cases, change the expression to a surd (an old-fashioned word for radical).

RJL Maplesoft

The OP mentions "...I can't get the integral to work with the dy." This leads me to suspect that the OP, apparently new to this forum and probably to Maple, does not realize that the Int command does not require a user to input the differential dy. The syntax for an unevaluated integral is Int(f(y),y=a..b). Maple really does not have a concept of a "differential".

If the integral defining BesselJ(0,x) is implemented as an inert integral by means of the Int command, then it can be evaluated by applying the value command.

RJL Maplesoft

I agree that without a global switch to see which text-field equations are "live" it can be tedious to hover over each one to check.

Prior to Maple 2016, one would exit typeset math by pressing Function Key F5. Now, you have to train yourself to press Shft-F5 to make the typeset math inert, then press F5 again to exit the typeset math field.

I have asked GUI to fix this so that Shift-F5 both makes the typeset math inert, and simultaneously exits the field. We'll have to see if the next version of Maple makes this more convenient.

RJL Maplesoft

Looks to me like there are two problems in the last line where delta is computed. The inner sum needs its range inside the parenttheses, and the second sum needs a range, not just the equation j=i. The first problem is easily corrected, but the second one appears to be structural. You might need to rethink what you are trying to calculate in delta.


There are tools for this in the Student MultivariateCalculus package.





Now all of these operations are also possible via the Context Menu system after the package has been installed. Also, the form of the line is not restricted to the one chosen here. See the help page for GetRepresentation.

@Annonymouse  Add in a graph of the intersection drawn by the intersectplot command from the plots package.


There is no value declared for x[0]. Add such a line to the first batch of initializations and the code runs.

If w is real and z complex, this is the root-locus problem of control theory. As noted by other contributors, there is no way to solve a sixth-degree polynomial equation in closed form, so obtaining exact (analytic) solutions for z=z(w) is a challenge. Depending on what you need z(w) for, some of the following devices might be of help.

Although Maple's plots package contains a rootlocus command, it numerically solves and graphs solutions for 1+k*h(z), where k is a real parameter (called a "gain" in the language of controls).

If z is real, then a graph of z=z(w) can be obtained by solving for w=w(z), then graphing via plot([w(z),z,z=a..b]).

If z is complex, solve for w=w(z) and replace z with x+I*y. Let Wr be the real part and Wi be the imaginary part. (This takes evalc(Re(... and evalc(Im(... but for polynomials, Maple can do this.) Now, if w is real, then Wi is zero, so an implicitplot of Wi=0 is a graph of the root locus. Each point on the resulting graph is a z=x+I*y value in the xy-plane for which some w satisfies the original equation P=0. The trick now is to obtain that w-value for each point on the curve.

For example, pick an x-value and count how many y-values will correspond. Then set x in Wr equal to that value and fsolve for y. Make a procedure of this process and you will be able to see the values of w for each z on the root locus. In control theory, a value of z indicates a possible state of a system. The idea is to determine the gain that will put the system into that state, so knowing the gain as a function of the state is a useful thing, something not ordinarily discussed in elementary texts in the subject.

As an isolated mathematical problem, I have always found this topic interesting. Just what is the trajectory in the z-plane of the solutions of P=0 as w varies, even if w isn't real. It's a hard problem to solve directly, so looking at it "backwards" certainly leads to interesting options.

The help page for pdsolve,numeric clearly states that the pdsolve command accepts a single, or a set, or a list, of time-dependent PDEs in two independent variables. Time is one of the independent variables, so there can be at most one spatial variable. Unfortunately, the situation is the same in Maple 2016.

RJL Maplesoft



In Maple 2016, open a help page and click the rightmost button appearing here under the Help menu. This toggles the examples between typeset input and text input. Not sure how the toggle looks, or where it's located in earlier versions, but the toggle has been available for a number of releases.


If you make the assignment A:=(1,2,3) and then ask Maple what is A, you find that A is just the sequence 1,2,3. In other words, round parentheses are for function evaluation, or for algebraic grouping. The construction (1,2,3) by itself has no meaning in Maple. That's why the Context Menu does not recognize it. It would recognize the sequence 1,2,3, or the set {1,2,3} or the list [1,2,3].

There is a way to trick Maple into including the round parentheses: write A=``(1,2,3) and the Context Menu will provide the option Assign Name. But be careful with this device. At some point either Maple will become confused as to the meaning of the object, or will simply strip the parentheses and treat the object as a sequence.

The equilibrium points of a system such as the one exhibited are solutions of the equations x'=y'=0. In this case, the graphical approach of plots:-implicitplot([sin(x)+y=0,y^2-x=0],x=-5..5,y=-5..5) willl show that the only real solution is (0,0). Alternatively, a bit of algebra leads to sin(y^2)=-y, and a graph of sin(y^2) and -y will suggest the same result.

Analytically, the solve command applied to the equations results in complicated expressions and complex solutions, along with the single real solution (0,0).

Maple has other tools for solving these equations; a proof that there is only one real solution is probably beyond the scope of the original question.

First 7 8 9 10 11 12 13 Last Page 9 of 22