area between 4sqrt(x), 2x; volume of revolution

[dmdtaz]

Can tell me if this is right.
y = 4square root(x) and y =2x S means intergal
1. a. Find the area of the region upper limit if figure it right is 7 lower is 0
A =S [(4x)^1/2 -2x)]
4/2(x)^(-1/2) -2x)]
4/2 (7)^(-1/2) -2(7) - 0
=13.24

b. find the V of the solid generated by revolving the region about the x axis. V= Pie S[f(x)]^2 dx upper limit b lower limit a
upper limit if I figure it out right is 4 lower limit is 0
Pie S (4square root(x)^2 -2(x)2
Pie 8x-4x2
pie (8)(4)
-4(4)^2
= 100.53

 

[skeeter]

Re: area volume

Can tell me if this is right.
y = 4square root(x) and y =2x S means intergal
1. a. Find the area of the region upper limit if figure it right is 7 lower is 0
A =S [(4x)^1/2 -2x)]
4/2(x)^(-1/2) -2x)]
4/2 (7)^(-1/2) -2(7) - 0
=13.24

to find the region enclosed by the two graphs, your upper limit of integration should be 4, not 7.

b. find the V of the solid generated by revolving the region about the x axis. V= Pie S[f(x)]^2 dx upper limit b lower limit a
upper limit if I figure it out right is 4 lower limit is 0
Pie S (4square root(x)^2 -2(x)2
Pie 8x-4x2
pie (8)(4)
-4(4)^2
= 100.53
\(\displaystyle V = \pi \int_0^4 16x - 4x^2 \, dx\)

btw ... the number represented by the greek letter is spelled pi, not pie.