Definite integral of functions of two variables (1 Viewer)

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1. The problem statement, all variables and given/known data
A pile of earth standing on flat ground near an abandoned mine has height 13 meters. The ground is the xy-plane; the origin is directly below the top of the pile and the z-axis is upward. The cross-section at height z is given by [tex]x^2 + y^2 = 13 - z[/tex] for [tex]0 \leq z \leq 13[/tex] with x, y, and z in meters.

(a) What equation gives the edge of the base of the pile?
(b) What is the area of the base of the pile?

2. Relevant equations



3. The attempt at a solution
I really don't know what it means by the edge of the base of the pile, I know that after we get to part a. then the area is just the integral right?
 
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Dick

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The 'edge of the base of the pile' is the intersection of the pile with the x-y plane. I.e. z=0. The 'area' is the area enclosed by that curve. You can find the area by integrating, but it's a pretty simple curve. You may know a formula.
 
So the edge of the base pile is [tex]x^2 + y^2 = 13[/tex] and I just need to integrate this right.. do I integrate it over x and y? I never knew an integral rule with that formula.. I think.

If I do an integration with respect to x and y I have:

[tex] (x^3/3 - 13x) + (y^3/3-13y) [/tex]
 
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Dick

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It's the area between the two functions y=+/-sqrt(13-x^2). But don't you recognize the shape of the curve?
 
so I just integrate sqrt(13-x^2)
 

Dick

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You integrate 2*sqrt(13-x^2) if you want to. But I can think of easier ways to find the area of a circle.
 
and how is that? and where did you get the 2 * in front of the sqrt(.........)
 

Dick

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Because you have to integrate between y=-sqrt(13-x^2) and y=+sqrt(13-x^2). You don't know how to find the area of a circle?
 
Yes it is [tex] \pi r^2 [/tex] and the radius is [tex] \sqrt(13) [/tex] right? How come when you do the integral way you don't get a pi anywhere??
 

Dick

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Yes it is [tex] \pi r^2 [/tex] and the radius is [tex] \sqrt(13) [/tex] right? How come when you do the integral way you don't get a pi anywhere??
You do get a pi. You have to do a trig subsitution to do the integral.
 
so the area is just [tex] 13 \pi [/tex]
 
And so if there's another question:

What equation gives the cross-section of the pile with the plane z = 10?
The answer is x^2+y^2 = 3 and what is the area of this?
 

Dick

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Do it the same way as x^2+y^2=13.
 
and I can just use the area of a circle with radius sqrt(3) as well here to make life easier?
 

Dick

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Why would you think not?
 
Just clarifying, and what is A(z), the area of a horizontal cross-section at height z?

is it just z = - x^2 - y^2 - 13
 

HallsofIvy

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Just clarifying, and what is A(z), the area of a horizontal cross-section at height z?

is it just z = - x^2 - y^2 - 13
1. This is not a function of z so it can't be the area function you want.

2. Since the formula for the cone is [itex]x^2+ y^2= 13- z[/itex], you want the equation
of the circular boundary at height z, not a equation for z. In fact, [itex]x^2+ y^3=13-z[/itex] is precisely what you want!
 
why is it z-13-z? where did you get the extra -z from? z-13-z is just 13 right?
 

HallsofIvy

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That was a typo. I have edited it.
 
if I take the integral of this area then the units would be meters^3? as it would be the volume
 

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