Evaluate the integral inside domain V, where V is bounded by the planes

In summary, the integral is not working because the user is getting stuck with integrals that are too difficult to integrate forwards.
  • #1
Cloudless
15
0
1. Evaluate the integral

VxdV

inside domain V, where V is bounded by the planes x=0, y=x, z=0, and the surface x2+y2+z2=1



Answer given: 1/8 - √2/16 (which is NOT what I got.. :yuck:)
2. The attempt at a solution

Ok, it's a triple integral, I know this.

∫dx runs from 0 to 1

∫dy runs from -√1-x2 to x

∫dz runs from 0 to √1-x2-y2

So the order of integrals go ∫∫∫dz dy dx

Yeah... it's not working. I keep getting stuck with things like x(1-y2-x2)1/2 which I can't integrate onwards... :cry:
 
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  • #2
Hi Cloudless! :smile:
Cloudless said:
VxdV

inside domain V, where V is bounded by the planes x=0, y=x, z=0, and the surface x2+y2+z2=1

this is a vertically sliced wedge from θ = 45° to 90°, so it would be much easier to use spherical coordinates :wink:

of course, you can use x y z, but …

∫dx runs from 0 to 1

no

∫dy runs from -√1-x2 to x

no
 
  • #3
In the xy-plane, z= 0 so [itex]x^2+ y^2+ z^2= 1[/itex] projects to the quarter circle [itex]x^2+ y^2= 1[/itex]. For each x, y ranges for 0 up to [itex]y= \sqrt{1- x^2}[/itex], NOT [itex]y= -\sqrt{1- x^2}[/itex] because the bounding surface is above the xy-plane.
 
  • #4
I redid it in spherical coordinates. Much easier, but still missing a pi/2. I think the conclusion is the answer key was wrong.
HallsofIvy said:
In the xy-plane, z= 0 so [itex]x^2+ y^2+ z^2= 1[/itex] projects to the quarter circle [itex]x^2+ y^2= 1[/itex]. For each x, y ranges for 0 up to [itex]y= \sqrt{1- x^2}[/itex], NOT [itex]y= -\sqrt{1- x^2}[/itex] because the bounding surface is above the xy-plane.



The question was ambiguous in its bounded area. I thought it was from negative pi/2 to pi/4.
 
  • #5
Cloudless said:
I redid it in spherical coordinates. Much easier, but still missing a pi/2. I think the conclusion is the answer key was wrong.

If you want us to check your answer, you'll have to show us your calculations. :wink:
 

What is the definition of an integral?

An integral is a mathematical concept that represents the accumulation of a quantity, such as area or volume, over a given interval. It is typically denoted by the symbol ∫ and is used to calculate the total value of a function within a given range.

What is the difference between a definite and indefinite integral?

A definite integral has specific limits or boundaries, while an indefinite integral does not. This means that a definite integral will give a specific numerical value, while an indefinite integral will result in a function.

How can I determine the bounds of a given domain?

The bounds of a domain can be determined by identifying the planes or curves that form its boundaries. This can be done by graphing the function or by setting up and solving equations to find the points of intersection.

Can the integral inside a bounded domain be negative?

Yes, the value of an integral inside a bounded domain can be negative. This can occur when the function being integrated has values that are both positive and negative within the given range.

What is the significance of evaluating an integral inside a given domain?

Evaluating an integral inside a given domain allows for the calculation of important quantities, such as area or volume, which have real-world applications in fields such as physics, engineering, and economics. It also allows for the determination of the average value of a function within a given range.

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