Volume of a tetrahedron by Triple Integral

faiz4000
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Homework Statement


By using triple integral, find the volume of the tetrahedron bounded by the coordinate planes and the plane 2x+3y+2z=6.

Homework Equations



Volume= ∫vdv=∫∫∫dxdydz

The Attempt at a Solution



find intercepts of the plane on the axes,
x-intercept=3
y-intercept=2
z-intercept=3then i don't know how to get limits of integration in the formula
 
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faiz4000 said:

Homework Statement


By using triple integral, find the volume of the tetrahedron bounded by the coordinate planes and the plane 2x+3y+2z=6.


Homework Equations



Volume= ∫vdv=∫∫∫dxdydz

The Attempt at a Solution



find intercepts of the plane on the axes,
x-intercept=3
y-intercept=2
z-intercept=3


then i don't know how to get limits of integration in the formula

Draw a picture of the plane in the first octant by joining those 3 points. Then use that picture for the limits.
 
Yes i get that. If I were finding the area of in 2D, i would draw lines parallel to x or y-axis and find the curves between which they lie. These would be the limits of inner integral. then i find the lowest and highest value of the outer integral and that becomes the limits for it... but in 3D i would have to draw planes parallel to say xy plane...but don't know which curves they lie between.
 

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faiz4000 said:
Yes i get that. If I were finding the area of in 2D, i would draw lines parallel to x or y-axis and find the curves between which they lie. These would be the limits of inner integral. then i find the lowest and highest value of the outer integral and that becomes the limits for it... but in 3D i would have to draw planes parallel to say xy plane...but don't know which curves they lie between.

It's the same idea in 3D. If you are going to do the inner integral in the ##z## direction first, you go from ##z## on the bottom surface (the xy plane) to ##z## on the top surface (the plane). Once you have done that you will have an xy integral and you can look at the triangle in the xy plane for the limits, just as you would do in 2D.

If you wanted to go in the x direction first for some reason, you would go from x on the back surface to x on the front surface then look in the yz plane for the dydz limits.
 
got it thanx.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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