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March 2, 2015

March 2, 2015

Posted by **naseba** on Thursday, March 29, 2012 at 8:40am.

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.

- calc -
**Reiny**, Thursday, March 29, 2012 at 9:02ama)

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)(4^(3/2)

= (2/3)(8) = 16/3

b)

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

- calc -
**naseba**, Thursday, March 29, 2012 at 9:07amthanks

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