posted by naseba .
1. Let R be the region bounded by the x-axis, the graph of y=sqr(x) , and the line x=4 .
a. Find the area of the region R.
b. Find the value of h such that the vertical line x = h divides the region R into two regions of equal area.
c. Find the volume of the solid generated when R is revolved about the x -axis.
d. The vertical line x = k divides the region R into two regions such that when these two regions are revolved around the x – axis, they generate solids with equal volumes. Find the value of k.
Area of region from x=0 to x=4
= ∫x^(1/2) dx from x=0 to 4
= [ (2/3)x^(3/2) ] from x = 0 to 4
= (2/3)(8) = 16/3
so we want
∫x^(1/2) dx from x = 0 to h to have a value of 8/3
(2/3)h^(3/2) = 8/3
h^(3/2) = (8/3)(3/2) = 4
h = 4^(2/3) = 16^(1/3) or cuberoot(16) or appr 2.52
c) this next part can be done in the same way
first let's find the volume from x-0 to 4
V = π∫ x dx from 0 to 4
= [(1/2)πx^2 ] from 0 to 4
= (1/2)π(16) = 8π
d) set the integral equal to 4π and solve for k
I am sure you can handle the rest