surface area of the solid (column side)

Question

I made a problem But I'm stuck in solving .. :-( the problem is following.

Find the surface area of

the solid that lies under the paraboloid $z =x^2 + y^2$,

above the $xy$-plane,

and inside the cylinder $(x-1)^2 + y^2 = 1$.

$$ $$ The solid is composed to 3 surface,

1. base side $ $ 2. convex upper side $ $ 3. column side

I got the area 1, 2

But I'm in trouble "No. 3 column side"

How can I calculate area of column side? Could you give me some idea?

Thanks in advance.

Answer 1

The intersection of the cylinder with the $x,y$ plane can be parametrized by $x(t)=1+\cos t,\ y(t)=\sin t.$ Now since the circle has radius $1$ the element $dt$ has unit differential length, and the height above the point $(x(t),y(t))$ is

$$x(t)^2+y(t)^2=(1+\cos t)^2+ (\sin t)^2,$$ which integrated for $t$ from $0$ to $2 \pi$ gives the surface area of the "column side", if that means the part on the boundary of the cylinder and below the parabaloid.

Answer 2

Using polar coordinates is advantageous.Using symbols gives opportunity for physics dimensional checks for length and area.

$ r^2 = x^2 + y^2 = a z $

Displaced circle in x,y plane $ (x-b)^2 + y^2 = b^2 $

$(x,y) = r ( cos \theta, sin \theta) $

Simplify, discard r =0 double point at origin.

$ r = 2\, b \,cos\theta $

$ r^2 = 4 b^2 cos^2 \theta = a z $

$ z = ( 4 b^2/a) cos^2 \theta $ is the height of tower as function of $ \theta.$

Area= $ \int z b d \theta/2 $, $ \theta$ limits between $0,\pi $(why?)

Now you can take it from there.

You may find cutting cylinder and sphere resulting in intersecting curve of Viviani also as a similar exercise.

EDIT1

$ a \, z $ is nothing but $z$.

I introduced the constant $a$ for physical dimension checking in a balanced dimension equation. You can take $a =1 $ for the maths purpose.

If you do the exercise with Viviani curve you notice that the radius of the sphere and diameter of cylinder are both a, cylinder generator coinciding with sphere axis. a matter you may not appreciate if you did not use symbol $a$ in the first place.

Answer 3

I think that the surface of the column side can be divided into two parts, one lies in the half space of $\{y\geq 0\}$ (which will be called the first part in the following), and the other in the half space of $\{y<0\}.$ Since by symmetry the areas of the two parts are identical, it suffices to calculate the part of that in $\{y\geq 0\}.$ Write $z=x^2+y^2$ as \begin{gather*} y=\sqrt{z-x^2} \end{gather*} and calculate the projection $D$ of the first part of the column side over the plane $XOZ,$ we have \begin{gather*} D=\{(x,z)\in\mathbb{R}^2\mid 0\leq z\leq 2x, 0\leq x\leq 2\}. \end{gather*} Now estimate the double integral (which is the area of the total column side) \begin{gather*} A_{\text{column side}}=2\iint_{D}\sqrt{y_x^2+y_z^2+1}\,\,\rm{d} x\rm{d}z \end{gather*}