Finding volume using triple integration

The question is Find the volume of the region cut from the solid elliptical cylinder x2+4y2≤4 by the xy plane and the plane z=x+2

My code is

Can anyone tell where i went wrong and also please tell whether my limits are correct

[darova]

please attach the code

Answer 1

[Yash Shingavi]

This shall work :

clear
clc
syms x y z real
xa=-2;
xb=2;
ya=0+0*x;
yb=sqrt(4-x^2);
za=0+0*x;
zb=x+2;
I=int(int(int(1+0*z,z,za,zb),y,ya,yb),x,xa,xb)
viewSolidone(z,za,zb,y,ya,yb,x,xa,xb)

[DGM]

This gives the correct answer, but for the wrong reason. The ellipse area is doubled, but only half of it is being considered.

syms x y z real
xa=-2;
xb=2;
ya=0+0*x;
yb=sqrt(4-x^2);
za=0+0*x;
zb=x+2;
I=int(int(int(1+0*z,z,za,zb),y,ya,yb),x,xa,xb)

I = 

viewSolid(z,za,zb,y,ya,yb,x,xa,xb)
axis equal

Answer 2

[SHAIK IMRAN]

syms x y z 
xa=-2;
xb=2;
ya=0+0*x;
yb=sqrt(4-x^2)/2;
za=0+0*x;
zb=x+2;
I=int(int(int(1+0*z,z,za,zb),y,ya,yb),x,xa,xb)
viewSolidone(z,za,zb,y,ya,yb,x,xa,xb)

[DGM]

In this case, the ellipse geometry is correct, but since only half of it is being considered as before, this gives half of the correct answer.

syms x y z 
xa=-2;
xb=2;
ya=0+0*x;
yb=sqrt(4-x^2)/2;
za=0+0*x;
zb=x+2;
I=int(int(int(1+0*z,z,za,zb),y,ya,yb),x,xa,xb)

I = 

viewSolid(z,za,zb,y,ya,yb,x,xa,xb)
axis equal

Answer 3

[Nithish]

clear
clc
syms x y z
int(int(x+2,y,0+0*x,sqrt(4-x^2)),x,-2,2)
viewSolid(z,0+0*x+0*y,x+2,y,-sqrt(4-x^2),sqrt(4-x^2),x,-2,2

[DGM]

Again, this gives the correct answer, but for the wrong reason. The ellipse geometry is wrong, though both halves are being considered.

syms x y z
int(int(x+2,y,0+0*x,sqrt(4-x^2)),x,-2,2)

ans = 

viewSolid(z,0+0*x+0*y,x+2,y,-sqrt(4-x^2),sqrt(4-x^2),x,-2,2)
axis equal

The only reason that two of these answers are coincidentally right is that the major radius of the ellipse is 2, its aspect ratio is also 2, and the volume has symmetry. If all of these things weren't conveniently interchangeable (i.e. if the ellipse geometry were different), the answers would cease to be accidentally right. The half-curve is y = sqrt(4-x^2)/2, so consider twice its integral to account for symmetry.

syms x y z
int(int(x+2,y,0+0*x,sqrt(4-x^2)/2)*2,x,-2,2)

ans = 

viewSolid(z,0+0*x+0*y,x+2,y,-sqrt(4-x^2)/2,sqrt(4-x^2)/2,x,-2,2)
axis equal

.. though there are other things that would need to be considered for a more generalized calculation.

Considering the form of this answer compared to the others, it should also be fairly clear that inlining everything makes the code hard to read.