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Find the volume V of the described solid S. The base of S is the triangular region with...
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You need to notice that the base of the solid is a triangle that is bounded by x axis, y axis and the line `x = -y + 5` .
You need to integrate the function `x = -y + 5` with respect to y, using disk method, hence, you need to evaluate the area of cross section, such that:
`A = ((5 - y)(5 - y)*sin 60^o)/2 => A = (sqrt3)/4*(5 - y)^2`
You need to evaluate the volume such that:
`V = (sqrt3)/4*int_0^5 (5 - y)^2 dy`
You should come up with the substitution, such that:
`5 - y = t => -dy = dt => dy = -dt`
Changing the limits of integration and variables, yields:
`V = (-sqrt3)/4*int_5^0 (t)^2 dt => V = (sqrt3)/4*int_0^5 (t)^2 dt`
`V = (sqrt3)/4*t^3/3|_0^5`
Using the fundamental theorem of calculus, yields:
`V = (sqrt3)/4*(5^3/3 - 0^3/3) = > V = 125sqrt3/12`
Hence, evaluating the volume of solid S, under the given conditions, yields `V = 125sqrt3/12` .
Posted by sciencesolve on March 1, 2013 at 6:26 PM (Answer #1)
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