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Area of a plane enclosed within a cube

  • #1

Homework Statement


Find the area of the plane y=3x enclosed within the cube formed by the planes x=y=z= [itex]\pm[/itex] 5

Homework Equations





The Attempt at a Solution



Using the equation y=3x, I found two points (x,y)=(5,1.66) and (x,y)=(-5,-1.66), then by plugging these points in the distance formula, I got the magnitude of one dimension equal to 10.54. I feel that the size of the other dimension must also be 10.54, so the area should be 111.09. But the answer given in my book is 105.4. I've been visualizing and thinking for hours on end about this, but I can't pin down what mistake I'm making.
It's such a simple problem, and I'm really very disappointed at my inability to think clearly. :/

Please help me. :confused:
 

Answers and Replies

  • #2
haruspex
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You've done the hard part! In the z direction, the plane is vertical, so what is the extent of the plane within the cube?
 
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  • #3
You've done the hard part! In the z direction, the plane is vertical, so what is the extent of the plane within the cube?
Hello haruspex,

Since the plane given is the xy plane, I think the extent of the plane in the z direction should be 0? :confused:
 
  • #4
Dick
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Hello haruspex,

Since the plane given is the xy plane, I think the extent of the plane in the z direction should be 0? :confused:
I think you mean the two intersection points in the x-y plane are (x,y)=(5/3,5) and (x,y)=(-5/3,-5). Your intersection surface is a rectangle, so yes, one side is approximately 10.54. I don't know how you are visualizing the other side to be the same. Try again. So far, every thing is just in the x-y plane. z goes from -5 to 5. So??
 
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  • #5
I think you mean the two intersection points in the x-y plane are (x,y)=(5/3,5) and (x,y)=(-5/3,-5).
I think this is the core misunderstanding from my part. The way I see it, the y=3x plane is just a tilted version of the bottom/top face of the cube (these faces have no z direction, do they?). So, I imagine a horizontal plane inside the cube, and this is why I think there should be no z involved... :confused:

I am wrong, I just need you to help me see why.

Thank you so much!
 
  • #6
haruspex
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I think this is the core misunderstanding from my part. The way I see it, the y=3x plane is just a tilted version of the bottom/top face of the cube (these faces have no z direction, do they?). So, I imagine a horizontal plane inside the cube, and this is why I think there should be no z involved... :confused:
A plane defined by y=3x takes all z values independently of x and y. It is the line y=3x in the xy plane extended vertically through all z values.
 
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  • #7
A plane defined by y=3x takes all z values independently of x and y. It is the line y=3x in the xy plane extended vertically through all z values.
Ohhh! Wow, I feel so stupid now. It makes perfect sense.

Thank you so much, haruspex! I REALLY appreciate your valuable help! Thanks a ton! :smile:
 

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