how to display the 3d graph for this problem?

i was able to plot the graph in 2d but in 3d i am unable to plot the graph around the red line

(use this code and execute and see)

clc

clear all

syms x

f = tan(pi*x/4)

fL = [0 1]

yr = 0

iL = [0 1]

finverse(f)

Volume = pi*int((finverse(f))^2,iL(1),iL(2));

disp(['Volume is: ', num2str(double(Volume))])

fx = inline(vectorize(f));

xvals = linspace(fL(1),fL(2),201);

xvalsr = fliplr(xvals);

xivals = linspace(iL(1),iL(2),201);

xivalsr = fliplr(xivals);

xlim = [fL(1)-0.5, fL(2)+0.5];

ylim = fx(xlim);

figure(1)

subplot(2,1,1)

hold on

plot(xvals, fx(xvals),'-b','LineWidth',2)

plot([yr yr],[fL(1) fL(2)],'-r','LineWidth',2)

fill([xvals, xvalsr],[fx(xvals), ones(size(xvalsr))*yr],[0.8 0.8 0.8],'FaceAlpha',0.8)

legend('Function Plot','Axis of Rotation','Filled Region')

title('Function y=f(x) and Region')

xlabel('x-axis')

ylabel('y-axis')

subplot(2,1,2)

hold on

plot(xivals,fx(xivals),'-b','LineWidth',2)

plot(-xivals,fx(xivals),'-m','LineWidth',2)

plot([iL(1) iL(2)],[yr yr],'-r','LineWidth',2)

fill([xivals, xivalsr],[fx(xivals), ones(size(xivalsr))*yr], [0.8 0.8 0.8])

fill([-xivals, -xivalsr],[ones(size(xivals))*yr, fx(xivalsr)], [1 0.8 0.8])

title('Rotated Region in xy-Plane')

xlabel('x-axis')

ylabel('y-axis')

figure(2)

[X,Y,Z] = cylinder(xivals-yr,100);

hold on

Z = iL(1)+ Z.*(iL(2)-iL(1));

surf(Z,Y,X+yr,'EdgeColor','none','FaceColor','flat','FaceAlpha',0.6)

plot([yr yr],[iL(1) iL(2)],'-r','LineWidth',2)

title('Volume generated by revolving a curve about the y-axis')

xlabel('x-axis')

ylabel('y-axis')

zlabel('Z-axis')

view(-22,32)

the code needs correction after the figure(2) line only or maybe before too i may be wrong

can anyone help please?

Answer 1

[Star Strider]

Try it with these changes —

hs = surf(X,Y,Z,'EdgeColor','none','FaceColor','flat','FaceAlpha',0.6);

hs.ZData = hs.ZData-0.5;

hs.YData = hs.YData+0.5;

rotate(hs, [1 0 0], 90)

I already made them in the code. This just nudges the cone to where it should be, and then rotates it.

syms x  
f(x) = tan(pi*x/4) 

f(x) = 

fL = [0 1] 

fL = 1×2

0 1

yr = 0

yr = 0

iL = [0 1] 

iL = 1×2

0 1

finverse(f)

ans(x) = 

Volume = pi*int((finverse(f))^2,iL(1),iL(2)); 
disp(['Volume is: ', num2str(double(Volume))]) 

Volume is: 1.2492

fx(x) = f(x);                                           % Inline Is Not Good Programming Practise
xvals = linspace(fL(1),fL(2),201); 
xvalsr = fliplr(xvals); 
xivals = linspace(iL(1),iL(2),201); 
xivalsr = fliplr(xivals); 
xlim = [fL(1)-0.5, fL(2)+0.5]; 
ylim = fx(xlim); 
figure(1) 
subplot(2,1,1) 
hold on 
plot(xvals, fx(xvals),'-b','LineWidth',2) 
plot([yr yr],[fL(1) fL(2)],'-r','LineWidth',2) 
fill([xvals, xvalsr],[fx(xvals), ones(size(xvalsr))*yr],[0.8 0.8 0.8],'FaceAlpha',0.8) 
legend('Function Plot','Axis of Rotation','Filled Region') 
title('Function y=f(x) and Region') 
xlabel('x-axis') 
ylabel('y-axis') 
subplot(2,1,2) 
hold on 
plot(xivals,fx(xivals),'-b','LineWidth',2) 
plot(-xivals,fx(xivals),'-m','LineWidth',2) 
plot([iL(1) iL(2)],[yr yr],'-r','LineWidth',2) 
fill([xivals, xivalsr],[fx(xivals), ones(size(xivalsr))*yr], [0.8 0.8 0.8]) 
fill([-xivals, -xivalsr],[ones(size(xivals))*yr, fx(xivalsr)], [1 0.8 0.8]) 
title('Rotated Region in xy-Plane') 
xlabel('x-axis') 
ylabel('y-axis')

figure(2)  
[X,Y,Z] = cylinder(xivals-yr,100); 
hold on 
Z = iL(1)+ Z.*(iL(2)-iL(1)); 
hs = surf(X,Y,Z,'EdgeColor','none','FaceColor','flat','FaceAlpha',0.6);
hs.ZData = hs.ZData-0.5;
hs.YData = hs.YData+0.5;
rotate(hs, [1 0 0], 90)
plot([yr yr],[iL(1) iL(2)],'-r','LineWidth',2) 
title('Volume generated by revolving a curve about the y-axis') 
xlabel('x-axis') 
ylabel('y-axis') 
zlabel('Z-axis') 
view(-22,32)
% view(80,30)
grid on

.

[lakshmi sampath reddy Pulagum]

but, as you can see,yes thank you for rotating it but the from the fig (1) the second graph you can see that when that function is rotated about y axis the shape dosent match the 3d figure.can you fix it ?

[lakshmi sampath reddy Pulagum]

there must be a u-curve cut in it.

[Star Strider]

‘there must be a u-curve cut in it.’

I have absolutely no idea what that means.

I took a guess as to what the desired result was. If I guessed in error, I will delete my Answer in a few hours, since it will have turned out to not be an Answer after all.

Not the first time for me, and and in all likelihood, definitely not the last.

.

[lakshmi sampath reddy Pulagum]

i have checked the points its just a function y=x rotated about y axis can you fix this whole thing please.

[lakshmi sampath reddy Pulagum]

it should be y=tan(pi*x/4) about y axis

[Star Strider]

I still have no idea what that means.

End of my day here. I will delete my Answer tomorrow morning.

[Star Strider]

Note that the plot, x is on the open interval . If I recall correctly, a cylinder (that would be used for this purpose) only has a positive radius, and in the y direction has values on the closed interval (for all current intents) so rotating the plot around any axis is going to be a problem. The function is an odd function (not symmetric about 0) so simply reflecting the positive portion (as could be done for a curve) is not an option. (This would the the situation for all odd functions, not only .)

How should a cylinder with a negative radius be created and plotted for this purpose?

.

[lakshmi sampath reddy Pulagum]

Oh i got it.I am really sorry to waste your precious time. :(.Thank you for answering :).

[Star Strider]

As always, my pleasure!

No worries, and my time was definitely not wasted in responding to the posted problem. I simply had a great deal of trouble understanding what the desired result is (and I am still not certain that I have).

.

Answer 2

[lakshmi sampath reddy Pulagum]

can you just rotate the function y=tan(pi*x/4) about y-axis or x=0 and plot in 3D.?

Answer 3

[lakshmi sampath reddy Pulagum]

I=imread('Capture.png')

imshow(I)

[lakshmi sampath reddy Pulagum]

see that graph and pls help me :).