# Thread: Two questions involving surface area.

1. ## Two questions involving surface area.

1. Find the area of that portion of the cylinder $x^2 + y^2 = 4y$ lying inside the sphere $x^2 + y^2 + z^2 = 16$ in the first octant.

2. Find the area of that portion of the surface $z^2 = x^2 + y^2$ above the first quadrant of the $xy$-plane for which $x < 1$ and $y < 2$.

I seriously have no idea how to do any of these.

2. Originally Posted by bleepbloop
1. Find the area of that portion of the cylinder $x^2 + y^2 = 4y$ lying inside the sphere $x^2 + y^2 + z^2 = 16$ in the first octant.

2. Find the area of that portion of the surface $z^2 = x^2 + y^2$ above the first quadrant of the $xy$-plane for which $x < 1$ and $y < 2$.

I seriously have no idea how to do any of these.
Are you familar with the surface area formula

$SA = \iint_R \sqrt{1 + f_x^2 + f_y^2}\, dA$

and with double integrals in polar coordinates?

3. Originally Posted by danny arrigo
Are you familar with the surface area formula

$SA = \iint_R \sqrt{1 + f_x^2 + f_y^2}\, dA$

and with double integrals in polar coordinates?
I know polar coordinates and double integrals and stuff, but I was never tought anything about the area of a 3D surface.

The only thing I got is

$ndS = \frac{\partial r}{\partial p}$x $\frac{\partial r}{\partial q}dpdq$

but I have no idea how to use it. The formula was given to me without any explanation.

4. Originally Posted by bleepbloop
I know polar coordinates and double integrals and stuff, but I was never tought anything about the area of a 3D surface.

The only thing I got is

$ndS = \frac{\partial r}{\partial p}$x $\frac{\partial r}{\partial q}dpdq$

but I have no idea how to use it. The formula was given to me without any explanation.
I think what you're refering to is the following.

If your surface $S$ is given parametrically as

$\vec{r} (p,q) = $ for $a < p < b,\;\;\;c < q < d$

then the surface area of S is given by

$\int_c^d \int_a^b || \vec{r}_u \times \vec{r}_v || dp\,dq$

where

$\vec{r}_p = < x_p, y_p, z_p>,\;\;\;\vec{r}_q = < x_q, y_q, z_q>,$

I think in the beginning it's easier to use cartesian coordinates and then switch to polar (if necessary) then trying to use this formula in general. If so, then

$\vec{r} = $ (here $p = x, \;\;q = y$)

then

$\vec{r}_x = <1, 0, f_x>$ and $\vec{r}_y = <0,1, f_y>$

so

$|| \vec{r}_x \times \vec{r}_y|| = \sqrt{1+f_x^2+f_y^2}$

Hence the formula I gave earlier. Are you familar with this

$\iint_R \sqrt{1+f_x^2+f_y^2} \,dx\,dy$?

5. Originally Posted by danny arrigo
I think what you're refering to is the following.

If your surface $S$ is given parametrically as

$\vec{r} (p,q) = $ for $a < p < b,\;\;\;c < q < d$

then the surface area of S is given by

$\int_c^d \int_a^b || \vec{r}_u \times \vec{r}_v || dp\,dq$

where

$\vec{r}_p = < x_p, y_p, z_p>,\;\;\;\vec{r}_q = < x_q, y_q, z_q>,$

I think in the beginning it's easier to use cartesian coordinates and then switch to polar (if necessary) then trying to use this formula in general. If so, then

$\vec{r} = $ (here $p = x, \;\;q = y$)

then

$\vec{r}_x = <1, 0, f_x>$ and $\vec{r}_y = <0,1, f_y>$

so

$|| \vec{r}_x \times \vec{r}_y|| = \sqrt{1+f_x^2+f_y^2}$

Hence the formula I gave earlier. Are you familar with this

$\iint_R \sqrt{1+f_x^2+f_y^2} \,dx\,dy$?
I've never seen that exact formula, but since it's derived from the formular I got, I guess that's what I have to do. Thanks.

6. I'll set you up on the first one - you try the second.

Here $z = f(x,y) = \sqrt{16-x^2-y^2}$ so $f_x = - \frac{x}{\sqrt{16-x^2-y^2}}$ and $f_y = - \frac{y}{\sqrt{16-x^2-y^2}}$ and

$\sqrt{1 + f_x^2 + f_y^2} = \sqrt{ 1 + \frac{x^2}{16-x^2-y^2} + \frac{y^2}{16-x^2-y^2}} = \frac{4}{\sqrt{16-x^2-y^2}}$

Now your region of integration is a circle $x^2+y^2 = 4y$ so we switch to polar coords., i.e, $r = 4 \sin \theta$ (the circle) and

$\sqrt{1 + f_x^2 + f_y^2} = \frac{4}{\sqrt{16-x^2-y^2}} = \frac{4}{\sqrt{16-r^2}}$.

For the limits of integration, $r = 0 \to 4 \sin \theta$ and $\theta = 0 \to \pi$. So the surface area is given by

$\int_0 ^{\pi} \int_0^{4 \sin \theta} \frac{4}{\sqrt{16-r^2}}\, r \, dr \, d \theta$

which you should be able to evaluate.

7. Originally Posted by bleepbloop
1. Find the area of that portion of the cylinder $x^2 + y^2 = 4y$ lying inside the sphere $x^2 + y^2 + z^2 = 16$ in the first octant.

2. Find the area of that portion of the surface $z^2 = x^2 + y^2$ above the first quadrant of the $xy$-plane for which $x < 1$ and $y < 2$.

I seriously have no idea how to do any of these.
The second problem is straight-forward. Simply use the formula given by the fellow above me and integrate along the bounds given. The first, however, is trickier because z is not a function of x,y. Cylinders are the three dimensional versions of vertical lines. That is, for a given x,y pair, there are an infinite number of z's. This type of problem can be solved in one of two ways.

1. You can rewrite the problem is terms of a dependent variable. Notice that while z is not a function of a given x,y pair, x is a function of a given z,y pair and y is a function of a given x,z pair. (If you restrict the range. Think of a circle in two dimensions y=+/-(1-x^2)^.5 ) This approach is somewhat difficult because you must discern this function.

2. Use vectors. This is much easier. Express the cylinder in terms of a vector valued function in two parameters. This way, the surface is a function relatively easily derived.

8. Originally Posted by danny arrigo
I'll set you up on the first one - you try the second.

Here $z = f(x,y) = \sqrt{16-x^2-y^2}$ so $f_x = - \frac{x}{\sqrt{16-x^2-y^2}}$ and $f_y = - \frac{y}{\sqrt{16-x^2-y^2}}$ and

$\sqrt{1 + f_x^2 + f_y^2} = \sqrt{ 1 + \frac{x^2}{16-x^2-y^2} + \frac{y^2}{16-x^2-y^2}} = \frac{4}{\sqrt{16-x^2-y^2}}$

Now your region of integration is a circle $x^2+y^2 = 4y$ so we switch to polar coords., i.e, $r = 4 \sin \theta$ (the circle) and

$\sqrt{1 + f_x^2 + f_y^2} = \frac{4}{\sqrt{16-x^2-y^2}} = \frac{4}{\sqrt{16-r^2}}$.

For the limits of integration, $r = 0 \to 4 \sin \theta$ and $\theta = 0 \to \pi$. So the surface area is given by

$\int_0 ^{\pi} \int_0^{4 \sin \theta} \frac{4}{\sqrt{16-r^2}}\, r \, dr \, d \theta$

which you should be able to evaluate.
The sphere delineates the boundary, not the surface to be calculated. The surface to be calculated is the cylinder. I believe that the problem was to determine the surface area of the cylinder inside the sphere, not the sphere over the cylinder. Unless I'm missing something.

9. Originally Posted by Bilbo Baggins
The sphere delineates the boundary, not the surface to be calculated. The surface to be calculated is the cylinder. I believe that the problem was to determine the surface area of the cylinder inside the sphere, not the sphere over the cylinder. Unless I'm missing something.
You're absolutely right! I did the reverse problem - the surface area of the sphere.

10. Originally Posted by danny arrigo
You're absolutely right! I did the reverse problem - the surface area of the sphere.
No biggie. Although, that fellow has logged off presumedly with the incorrect answer. However, I figured it out. (I think) Check it out.

This was challenging. The parametric equation for the cylinder was complicated by its intersection with the sphere, but I managed it. Ok Check it.
On the z = 0 plane, x and y trace out a circle with radius two centered at x=0 y=2 Parametrically, that can be expressed as (2*cos u) i + (2 + 2*sin u) j Now, z is given by +/- (16-x^2-y^2)^.5 We'll work exclusively with the positive half and just double our answer because everything is symmetric with regard to z=0. Further, the range of z is constrained within the restricted domain of x and y. That is, x^2+(y-2)^2 must equal 4. That is after all, the cylinder that we are interested in. So, we are looking for the Z values that lie on the surface of the sphere and lie directly above that circle drawn out on the z=0 plane. We already have an equation for y given by the cylinder. That is, y= +/- (4-x^2)^.5 + 2 By substituting this equation into the equation for the top hemisphere of the sphere we get z in terms of x alone. Z=(16+/-4*(4-x^2)^.5 -8)^.5 Now, we already expressed x in terms of a parameter u above. So, the equation becomes Z= (16+/-4*(4-4*(cosu)^2)^.5 -8)^.5 Now, we have a parametric equation for a curve in space, but not a surface. For that, we introduce a second parameter v and apply it to the K vector. Further, we allow v to run from 0 to 1 and no further. So finally we have the parametric equation:

R(u,v) = (2 cosu)i + (2+2sinu)j + v* ((16+/-4*(4-4*(cosu)^2)^.5 -8)^.5)k

To determine the surface area, use the cross product of the two partial derivatives.

11. Originally Posted by Bilbo Baggins
No biggie. Although, that fellow has logged off presumedly with the incorrect answer. However, I figured it out. (I think) Check it out.

This was challenging. The parametric equation for the cylinder was complicated by its intersection with the sphere, but I managed it. Ok Check it.
On the z = 0 plane, x and y trace out a circle with radius two centered at x=0 y=2 Parametrically, that can be expressed as (2*cos u) i + (2 + 2*sin u) j Now, z is given by +/- (16-x^2-y^2)^.5 We'll work exclusively with the positive half and just double our answer because everything is symmetric with regard to z=0. Further, the range of z is constrained within the restricted domain of x and y. That is, x^2+(y-2)^2 must equal 4. That is after all, the cylinder that we are interested in. So, we are looking for the Z values that lie on the surface of the sphere and lie directly above that circle drawn out on the z=0 plane. We already have an equation for y given by the cylinder. That is, y= +/- (4-x^2)^.5 + 2 By substituting this equation into the equation for the top hemisphere of the sphere we get z in terms of x alone. Z=(16+/-4*(4-x^2)^.5 -8)^.5 Now, we already expressed x in terms of a parameter u above. So, the equation becomes Z= (16+/-4*(4-4*(cosu)^2)^.5 -8)^.5 Now, we have a parametric equation for a curve in space, but not a surface. For that, we introduce a second parameter v and apply it to the K vector. Further, we allow v to run from 0 to 1 and no further. So finally we have the parametric equation:

R(u,v) = (2 cosu)i + (2+2sinu)j + v* ((16+/-4*(4-4*(cosu)^2)^.5 -8)^.5)k

To determine the surface area, use the cross product of the two partial derivatives.
Wow, thanks a lot. This number is so effed up. But #2 was pretty easy after danny arrigo's help.

Just one question: what are my integration boundaries gonna be? 0 -> 2pi for u and 0 -> 1 for v?

12. Originally Posted by bleepbloop
Wow, thanks a lot. This number is so effed up. But #2 was pretty easy after danny arrigo's help.

Just one question: what are my integration boundaries gonna be? 0 -> 2pi for u and 0 -> 1 for v?
Yes. Remember, you need to double your answer because there's a second half of the cylinder beneath the z=0 plane.

Now that I look at the expression for K again, I realize that it can be simplified significantly. It simplifies to v* 2*(2 +/- 2*sinu)^.5 After you determine the magnitude of the cross product of the two partial derivatives, the final integral that needs to be solved is just 4(2 +/- 2 sinu)^.5 dudv 2pi-0 and 1-0 respectively. I don't know how the +/- works out. I've never attempted it before.

Now that I think about it, are you sure that you have the problem correct? This seems a bit challenging for someone just learning surface area. Do you think that you mistransposed the problem? Integrating the sphere over the cylinder is MUCH easier and seems more like a beginer's problem than the cylinder within the sphere.

13. Originally Posted by Bilbo Baggins
No biggie. Although, that fellow has logged off presumedly with the incorrect answer. However, I figured it out. (I think) Check it out.

This was challenging. The parametric equation for the cylinder was complicated by its intersection with the sphere, but I managed it. Ok Check it.
On the z = 0 plane, x and y trace out a circle with radius two centered at x=0 y=2 Parametrically, that can be expressed as (2*cos u) i + (2 + 2*sin u) j Now, z is given by +/- (16-x^2-y^2)^.5 We'll work exclusively with the positive half and just double our answer because everything is symmetric with regard to z=0. Further, the range of z is constrained within the restricted domain of x and y. That is, x^2+(y-2)^2 must equal 4. That is after all, the cylinder that we are interested in. So, we are looking for the Z values that lie on the surface of the sphere and lie directly above that circle drawn out on the z=0 plane. We already have an equation for y given by the cylinder. That is, y= +/- (4-x^2)^.5 + 2 By substituting this equation into the equation for the top hemisphere of the sphere we get z in terms of x alone. Z=(16+/-4*(4-x^2)^.5 -8)^.5 Now, we already expressed x in terms of a parameter u above. So, the equation becomes Z= (16+/-4*(4-4*(cosu)^2)^.5 -8)^.5 Now, we have a parametric equation for a curve in space, but not a surface. For that, we introduce a second parameter v and apply it to the K vector. Further, we allow v to run from 0 to 1 and no further. So finally we have the parametric equation:

R(u,v) = (2 cosu)i + (2+2sinu)j + v* ((16+/-4*(4-4*(cosu)^2)^.5 -8)^.5)k

To determine the surface area, use the cross product of the two partial derivatives.
Although choice for R is

$\vec{r} (u,v) = < 2 \sin 2u,\; 2 + 2 \cos 2u,\; 4v \cos u\, >$, $0 \le u \le \pi,\;\;\; 0 \le v \le 1$

My choice of $x\; \text{and} \;y$ allow one to simplify $z$ in the parametric surface and further,

$||\, \vec{r}_u \, \times\, \vec{r}_v \, || = 4 \sin u$,

something rather easy to integrate.

14. Originally Posted by danny arrigo
Although choice for R is

$\vec{r} (u,v) = < 2 \sin 2u,\; 2 + 2 \cos 2u,\; 4v \cos u\, >$, $0 \le u \le \pi,\;\;\; 0 \le v \le 1$

My choice of $x\; \text{and} \;y$ allow one to simplify $z$ in the parametric surface and further,

$||\, \vec{r}_u \, \times\, \vec{r}_v \, || = 4 \sin u$,

something rather easy to integrate.
Hi. Are they the same? How did you derive this formula? I must admit, it looks far less foreboding than mine. However, I think that something may be incorrect here. (No offense) When u=pi and v=1, in your parametric, I get x=0 y=4 and z= -4 This point lies outside the sphere, and obviously, outside the portion of the cylinder in which we are interested. This is easily confirmed by plugging the values of x=0 y=4 and z=-4 into the equation of the sphere. Clearly, they don't fit. When x=0 and y=4, we are at the "apex" of the circle on the y axis. At this point, the sphere and cylinder meet. Z can have only value here, 0. Let me know what you think.

15. OK. I see what you've done. You have given the parametric for the cylinder, however, you've neglected the boundary. Your parametric cylinder is a complete cylinder that runs from z=4 to z= -4 You must realize that he originally asked for the surface area of the portion of this cylinder lying only within the given sphere. Think of what the intersection looks like. It's complicated. We're not talking about a plane here. The intersection with a sphere is going to be complex.

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