Area of a plane enclosed within a cube

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SUMMARY

The discussion focuses on calculating the area of the plane defined by the equation y=3x, which is enclosed within a cube formed by the planes x=y=z=±5. The user initially miscalculated the area by assuming both dimensions were equal, leading to an incorrect area of 111.09 instead of the correct area of 105.4. The key misunderstanding was regarding the vertical extent of the plane, which extends through all z-values, thus affecting the area calculation.

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Homework Statement


Find the area of the plane y=3x enclosed within the cube formed by the planes x=y=z= \pm 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:
 
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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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haruspex said:
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:
 
ViolentCorpse said:
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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Dick said:
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!
 
ViolentCorpse said:
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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haruspex said:
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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