Let f (y) = e^-3y/2. Let R be the region between the graph of f and the y axis o
ID: 2894786 • Letter: L
Question
Let f (y) = e^-3y/2. Let R be the region between the graph of f and the y axis on the interval [0, 1] (on the y axis). Set up an integral to find the volume V of the solid region obtained by revolving R about the y axis and then find the volume. V = integral^1 _0 = (exact value) Find the value of V to five decimal places: V = Let f (x) = 6x and g (x) = x^2 + 3x, and let R be the bounded region between the graphs of f and g. Set up an integral to find the volume V of the solid generated by revolving R about the x axis and then find V. V = integral =Explanation / Answer
1)
setup: V=[0 to 1]((e-3y/2)2)dy
V=[0 to 1]e-3ydy
V=[0 to 1](-1/3)e-3y
V=(-1/3)[e-3*1-e-3*0]
V=(-1/3)[e-3-e0]
V=(-1/3)[e-3-1]
V=(/3)(1-e-3)----------------->exact value
V=0.99506
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2)
when curves intersect 6x=x2+3x
=>x2+3x-6x=0
=>x2-3x=0
=>x(x-3)=0
=>x=0,x=3
volume,V=[0 to 3]((6x)2-(x2+3x)2) dx
V=[0 to 3](36x2-x4-6x3-9x2) dx
V=[0 to 3](-x4-6x3+27x2) dx
V=[0 to 3](-(1/5)x5-(3/2)x4+9x3)
V=(-(1/5)35-(3/2)34+9*33) -(-(1/5)05-(3/2)04+9*03)
V=(-(243/5)-(243/2)+243) -0
V=729/10
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