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Iterated integral clarification

  • Thread starter Derill03
  • Start date
There is an iterated integral:

outer integration = 1 to -1 (limits of int.)
inner integration = sqrt(1-x^2) to -sqrt(1-x^2)
function = x^2 + y^2

the part i dont get is the question states to sketch or describe the region, im good with that, but i dont understand what is meant by the following:

What region in the plane comprises the lower boundary of this region?

Can anyone help?
 

tiny-tim

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… i dont understand what is meant by the following:

What region in the plane comprises the lower boundary of this region?
Hi Derill03! :smile:

I agree it's a strange question :confused:

the boundary of a region in the plane is a curve, not a region …

I suppose it's asking for the bottom half of the circle.
 
Hello Derill03

I believe I am faced with the same question as you.

My question is what the graph of this would look like in R3. Is it an inverted conical shape or a sphere?

The lower boundary that TinyTim was mentioning, "lower half of the circle", is this referring a lower half of a sphere?
 
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Hello Derill03

I believe I am faced with the same question as you.

My question is what the graph of this would look like in R3. Is it an inverted conical shape or a sphere?

The lower boundary that TinyTim was mentioning, "lower half of the circle", is this referring a lower half of a sphere?
No, it's the lower half of a circle in the x-y plane. We're talking about the region over which (double) integration takes place, so this region is two-dimensional. OTOH, the integral itself might represent the volume of a 3D object whose z-value is x^2 + y^2, which is a cone.

Derill03,
You listed the limits of integration as
outer integration = 1 to -1 (limits of int.)
inner integration = sqrt(1-x^2) to -sqrt(1-x^2)
The usual practice is to list the lower limit first, and then the upper limit, not the other way round, as you seem to have done.
 

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