Let T be the tetrahedron with vertices (0 , 0 , 0), ( a, 0 , 0), (0 ,b, 0), and
ID: 2834357 • Letter: L
Question
Let T be the tetrahedron with vertices (0,0,0), (a,0,0), (0,b,0), and (0,0,c),
where a, b, and c are all positive.
(a) Set up an appropriate double integral for the volume of T .
(b) Set up an appropriate triple integral for the volume of T .
(c) Compute one of your integrals in (a) or (b).
(d) Check your answer in (c) by using the formula V = (1/3)Bh, where B is the area of the base
of T and h is the height of T.
Let T be the tetrahedron with vertices (0,0,0), (a,0,0), (0,b,0), and (0,0,c), where a, b, and c are all positive. Set up an appropriate double integral for the volume of T . Set up an appropriate triple integral for the volume of T . Compute one of your integrals in (a) or (b). Check your answer in (c) by using the formula V = (1/3)Bh, where B is the area of the base of T and h is the height of T.Explanation / Answer
The plane containing the three vertices is given by x/a + y/b + z/c = 1.
Solving for z:
z = c(1 - x/a - y/b).
Projecting T onto the xy-plane gives the region bounded between x/a + y/b = 1 and the x,y axes.
==> y = 0 to y = b(1 - x/a) with x in [0, a].
(a) The volume equals ?(x = 0 to a) ?(y = 0 to b(1 - x/a)) c(1 - x/a - y/b) dy dx
(b) The volume ??? 1 dV equals
?(x = 0 to a) ?(y = 0 to b(1 - x/a)) ?(z = 0 to c(1 - x/a - y/b)) 1 dz dy dx
(c) taking the double integral
The volume equals ?(x = 0 to a) ?(y = 0 to b(1 - x/a)) c(1 - x/a - y/b) dy dx
= ?(x = 0 to a) c[(1 - x/a)y - (1/b)y^2/2] {for y = 0 to b(1 - x/a)} dx
= ?(x = 0 to a) (bc/2) (1 - x/a)^2 dx
= ?(x = 0 to a) (bc/2) * (-a/3)(1 - x/a)^3 {for x = 0 to a}
= abc/6.
(d) The area of the base (bounded by y = 0 to y = b(1 - x/a) with x in [0, a])
equals (1/2)ab, being a triangle.
So, V = (Ah)/3 = (1/2)ab * c/3 = abc/6.
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