# How would you find the volume of the tetrahedron T bounded by the coordinate planes and the plane 3x+4y+z=10?

Volume = 125/9 (

The coordinate planes are given by

The given tetrahedron,

The line joining

# y-0 = ((5/2)/(-10/3))(x-10/3) #

# :. y = -3/4x+5/2 #

And so the region

# R = {(x, y) | 0 le x le 10/3, 0 le y le −3/4x +5/2 } #

And the volume,

integral of the function

# :. V = int int_R (10 − 3x − 4y) \ dA#

# " "= int_0^(10/3) int_(0)^(-3/4x+5/2) (10 − 3x − 4y) \ dy \ dx#

# " "= int_0^(10/3) [(10-3x)y-2y^2]_(y=0)^(y=-3/4x+5/2) \ dx #

# " "= int_0^(10/3) {(10-3x)((-3x)/4+5/2)-2((-3x)/4+5/2)^2} - {0} \ dx #

# " "= int_0^(10/3) (-30/4x+25+9/4x^2-15/2x)-2(9/16x^2-15/4x+25/4) \ dx #

# " "= int_0^(10/3) (-15/2x+25/2+9/8x^2) \ dx #

# " "= [-15/4x^2+25/2x+9/24x^3]_0^(10/3) #

# " "= {-15/4*100/9+25/2*10/3+3/8*1000/27} - {0} #

# " "= -125/3+125/3+125/9 #

# " "= 125/9 #

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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