Given the surfaces $S_1$, $S_2$, and $S_3$: $$ S_1:x^2+(y-1)^2=1 $$ $$ S_2:z=4-x^2-y^2 $$ $$ S_3:z=4 $$ Find the surface area of $S_1$ between $S_2$ and $S_3$.

My attempt:

The area of the surface would be $$ \iint_{S^*} 1 dS $$ where $S^*$ is the portion of the surface $S_1$ bounded by $S_2$ and $S_3$. But I can't figure out how to parametrize the surface. Looking from the top view, it's going to be like having 2 circles (one from the cross section of the cylinder $S_1$, and one from the cross section of the paraboloid $S_2$) intersecting each other, where the points of intersection depend on the height $z$. The intersection point (depending on $z$) can be found by solving these two equations $$ S_1:x^2+(y-1)^2=1 \quad \textrm{and} \quad S_2:z=4-x^2-y^2 $$ which it turns out to be $$ C : (x,y,z) = \left( \sqrt{(4-z)-\frac{(4-z)^2}{4}}, \frac{4-z}{2}, z \right) \quad z=[0,4] $$ Now I know where the paraboloid and the cylinder intersect each other, and I think I would know how to parametrize this surface too (with some algebra work). However, I anticipate that it's going to be too tedious to compute the surface area using this surface parameterization. So, is there any better way?

I was also thinking of using stokes' theorem. But I think the calculation is not going to be very nice as I need to parametrize the curve of the intersection $C^*$ by these two surfaces, and also need to find the nice function $F$ such that $$ \iint_{S^*} 1 dS = \iint_{S^*}(\nabla\times \vec{F})\cdot \vec{n} dS = \oint_{C^*} \vec{F}\cdot \vec{n} ds$$


1 Answer 1


It has to be noted that the surface cannot be projected onto $xy$-plane. We see that it is entirely in the positive $y$ portion, and symmetric. So we project it onto $yz$-plane, then multiply it by $2$. The following is the projection:

enter image description here

We want to find the part of $ABC$. The surface is represented by $x=\sqrt{2y-y^2}$. So

$$\iint_D \sqrt{1+(\frac{dx}{dy})^2+(\frac{dx}{dz})^2} dydz,$$ where $\frac{dx}{dz}=0, \frac{dx}{dy}=\frac{1-y}{\sqrt{2y-y^2}}$.

Now the limits for $z$ would be $4-y^2$ and $4$, since $x=0$, limits for $y$ are $0$ and $2$. It is not hard to get the following integral: $$\int^2_0 \frac{y^2}{2y-y^2}dy.$$

Here you will need to do partial fractions. Now you can try to continue.

  • $\begingroup$ Thank you, I was blind to the fact that it can be projected on to the yz-plane. $\endgroup$
    – IgNite
    Jun 10, 2016 at 2:48
  • $\begingroup$ @Ignite: you are welcome. Upon review, the last integral can be done by partial fractions. $\endgroup$
    – KittyL
    Jun 11, 2016 at 9:07

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