# Surface portion bounded by plane 2x +5y + z = 10 that lies

1. Oct 26, 2016

1. The problem statement, all variables and given/known data
Find the surface portion bounded by plane 2x +5y + z = 10 that lies in cylinder (x^2) +(y^2) = 9 ...

I have skteched out the diagram and my ans is 5sqrt(30) instead of 9sqrt (30) as given by the author ...
Anything wrong with my working ?

2. Relevant equations

3. The attempt at a solution

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2. Oct 26, 2016

### Ray Vickson

Your handwriting is so hard to read that I will not even try. Type out your work, as is the PF standard. (Read the post "Guidelines for students and helpers", by Vela.)

3. Oct 26, 2016

Which part you can't see it clearly?

4. Oct 26, 2016

### Staff: Mentor

Regarding your answer, I don't see that you have used the given information that the portion of the plane you want is inside the cylindrical cylinder.
The part between the top and bottom of what you wrote. I agree with Ray that your hand-written work is hard to read, and we strongly discourage posting the work in an image. Except for the sketches of the surface and the triangle, everything can be done using LaTeX. For a tutorial, see https://www.physicsforums.com/help/latexhelp/ (under the INFO menu, in the Help/How-to submenu)
Here's a quick lesson:
Fractions:
$\frac{a + b}{c}$ -- LaTeX script $\frac{a + b}{c}$
Square roots: $\sqrt{30}$ -- LaTeX script $\sqrt{30}$
Integrals $\int_{z = 0}^{2} f(z) dz$ -- LaTeX script $\int_{z = 0}^{2} f(z) dz$
$\int_{x = 0}^{10}\int_{y = 5}^{x} f(x, y) dy dx$ -- LaTeX script $\int_{x = 0}^{10}\int_{y = 5}^{x} f(x, y) dy dx$

5. Oct 26, 2016

### LCKurtz

@Mark44 What's with that LaTeX script thing? I like it and have had many occasions where I would have liked to use it, but it doesn't work for me, even when I copy yours as below. It just renders it:

LaTeX script $2^2$
LaTeX script $\int_{z = 0}^{2} f(z) dz$

6. Oct 26, 2016

### Staff: Mentor

I did something tricky in the part where I show the unrendered LaTeX: In the first # sign of the start and end pair, I change the text color, using the black and white circle (the fourth icon from the left on the menu bar. It doesn't matter what color you choose -- the presence of a bbcode color tag keeps the browser from rendering the LaTeX script as it normally would.

7. Oct 27, 2016

$\int_{y = 0}^{2} \int_{z=0}^{10-5y} \sqrt {(-0.5)^2 +(-2.5)^2 +1} dzdy$[/QUOTE]

$\int_{y = 0}^{2} \int_{z=0}^{10-5y} \ 0.5 sqrt {30} dzdy$

$0.5 sqrt {30}\int_{y = 0}^{2} \ dy$
=
$5sqrt {30}$

@Mark44 , here's my working , i am not sure whether my projection to zy plane is correct or not ...

8. Oct 28, 2016

anyone else can help ?

9. Oct 28, 2016

### LCKurtz

I have a couple of comments for you. First: your statement of the problem says the area of the plane inside the cylinder, but your picture and working apparently are for only the portion of the surface in the first octant only. Which is it?

Second: The intersection of a plane and a cylinder is typically an ellipse. Why on earth would you choose the projection on the zy plane? The only projection that would be circular would be on the xy plane. That's where you should project it.

10. Oct 28, 2016

Sorry , i forgot to include other octant ....
why the projection to xy plane is a circle ? why shouldnt the projection of xy plane is the area of the plane that lies inside the cylinder ? refer to the green part ... that's the part of plane

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11. Oct 28, 2016

### LCKurtz

No it isn't. The boundary of a plane cutting through a cylinder at an angle will be curved and, in this case, is an ellipse shape.

12. Oct 28, 2016

why ? can you explain further with the aid of diagram ? I still cant imagine it ....Btw , the plane 2x + 5y + z = 10 only lies in first octant , am i right ?

13. Oct 28, 2016

### LCKurtz

A plane extends in all directions. Your triangle is just a small portion of a plane. Forget the coordinate system for a minute. Just imagine a cylinder with a plane slicing through it. Here's a picture of one (not your particular one). There aren't any straight lines like you have drawn. Also, although the green is an ellipse, do you see that if you looked at it straight down the z axis it would look like a circle?

14. Oct 29, 2016

ok , i can undetstand that .... But , i still couldnt imagine that why my case is same as above (the cylinder that is being sliced above) ... The plane 2x+5y +z =10 is a triangular plane , right ? It's not the top part of cylinder that is being sliced

15. Oct 29, 2016

### LCKurtz

Here's what your original problem statement was:

There is nothing about triangles, octants, or straight lines. When planes cut through cylinders at an angle it looks generally like the picture I gave you.

16. Oct 29, 2016

why ? The 2x+5y +z =10 is a triangular plane... The diagram that you sketched look like the volume bounded by tirangular plane 2x+5y +z =10 and cylinder ... But , what we want here is the surface area of plane 2x+5y +z =10 inside cylinder , right ? So , how could the surface area of tirangular plane 2x+5y +z =10 look like a circular plane ?

17. Oct 29, 2016

### LCKurtz

I can't tell whether you are trolling me or are just hopeless. Either way it's pretty clear I can't get through to you with this problem so I am abandoning this thread. Maybe someone else will continue.

18. Oct 29, 2016

### Staff: Mentor

No!!!
A plane is not triangular nor is it circular, as you say below.
The diagram that LCKurtz provided shows the surface area (not volume) of the region of a plane inside a cylinder, very similar to what you're trying to find in this problem.

Until you can understand why the diagram that you drew is incorrect, and the one that LCKurtz provided is what you need, you will not be able to do this problem.
Again, your statements about a "triangular plane" and a "circular plane" make no sense.

19. Oct 29, 2016

if we look at the plane 2x+5y +z =10 only , then it's triangular plane right ?

20. Oct 29, 2016

### Staff: Mentor

No! If you look at only the portion of the plane in the first octant, what you get is a triangle, but the plane extends infinitely far in two dimensions.

BTW, there is no such thing as a "triangular plane" or a "circular plane."