# 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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Mukesh
on 23 Nov 2022

%Evaluation of Volume of solid of revolution

clear

clc

syms x

f(x)=4/(x^2+4); % Given function

yr=0; % Axis of revolution y=yr

I=[0,2]; % 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.

surf(Z,Y+yr,X) % Plotting the solid of revolution about y=yr

hold on;

plot([a b],[yr yr],'-r','LineWidth',2); % Plotting the line y=yr

view(22,11); % 3-D graph viewpoint specification

xlabel('X-axis');ylabel('Y-axis');zlabel('Z-axis');

### Answers (1)

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.

syms x

y(x) = 4/x^2 + 4

fplot(y,[0,2])

ylim([0,100])

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.

int(y,[0,2])

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.

int(pi*y^2,x,[0,2])

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)

fplot(y2,[0,2])

ylim([0,1])

Clearly the surface generated by revolving that curve around the x axis will be finite.

int(pi*y2^2,x,[0,2])

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