Find the volume of the solid generated by revolving about the x-axis the region bounded by the curve y= 4/x^2+4,the axis, and the lines x=0 x=2
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%Evaluation of Volume of solid of revolution clear clc syms x f(x)=sqrt(x); % Given function yr=1; % Axis of revolution y=yr I=[0,4]; % Interval of integration a=I(1);b=I(2); vol=pi*int((f(x)-yr)^2,a,b); disp('Volume of solid of revolution is: '); disp(vol); % Visualization if solid of revolution fx=matlabFunction(f); xv = linspace(a,b,101); % Creates 101 points from a to b [X,Y,Z] = cylinder(fx(xv)-yr); Z = a+Z.*(b-a); % Extending the default unit height of the %cylinder profile to the interval of integration.
John D'Errico on 21 Sep 2023
The question is now a year old, so I'll show how to solve it. @Mukesh chose to work with a different domain than the question asks in the title. While that seems surprising, the problem as posed in the subject title does not have a finite solution.
A surface of revolution of the function y(x), bounded by x==0 and x==2, revolving around the x axis.
y(x) = 4/x^2 + 4
I've chopped it off at y==100 there. The first question is to know if it is integrable at all. We can get a clue by this computation, merely finding the area under that curve. As you can see, the tail as x--> 0 is too heavy, so we would expect the volume to also be unbounded.
How would we compute the volume inside that surface of revolution? Thw simple way is to view this as a set of infinitessimally thin circles, with center at y==0, and with radius y(x). The area of each circle is just pi*y(x)^2. Then just integrate from x==0 to x==2.
And, as expected, we get an infinite volume. That suggests, that @Mukesh was possibly correct in the comment where the function y(x)=4/(x^2+4) was assumed, even though that is NOT mathematically what was written. (SIGH.) If we do that, the region of interest is now:
y2(x) = 4/(x^2+4)
Clearly the surface generated by revolving that curve around the x axis will be finite.