The region bounded by y=sqrt x, y=0, x=0, and x=4 is revolved about the x-axis.?

Question

find the value of x in the interval [0,4] that divides the solid into two parts of equal volume. could anyone show me the steps to solve this problem in detail.

Answer 1

Equate volume, using the unknown limit of integration:



Inegral {0 to k} [pi * (sqrt(x))^2] dx



=



Integral {k to 4} [pi * (sqrt(x))^2] dx



---->



Inegral {0 to k} [pi * x] dx

=

Integral {k to 4} [pi * x] dx



---->

(x^2)/2 {0 to k} = (x^2)/2 {k to 4}



[(k^2)/2 - 0] = [8 - (k^2)/2]

k^2 = 8

k = 2*sqrt(2)



so, x= 2*sqrt(2)

Answer 2

Hi.



This problem requires a bit of knowledge about integrals (there may be some tricky resolution without them, but I'll assume you are studying them if you faced this problem).



It is a bit hard to write integrals in simple HTML, but I hope you will be able to understend it. So let's go:



The first step is (always) to plot (or draw) a graphic for the region. It will give you an overview of what is asked by the problem.



In this case, you have a a picture *LIKE* this one (it is not *this* one, since yours is bounded by y=sqrt(x), this one is more like y=x^2, and yours goes from 0 to 4, and this one from 0 to 2, but it will serve as an illustration):



http://chuwm2.tripod.com/revolution/line2.gif



The rotation will make it look like the following one (again, yours will look a bit different - more like a cap than a cone, because the upper line is curved in the opposite direction):



http://chuwm2.tripod.com/revolution/solid3.gif



The important theorem that will help you is this one: "The volume of a solid created by the rotation of a function alongside the x axis, considering the region of the function between a and b is pi.integral from a to b of the square of the function", that is:



V = π.∫(f(x)²)dx (where the integral is from a to b)



In your case, you will divide your region (which ranges from 0 to 4) in two regions, forming two solids: a "tip" and a "base" (in the picture, you would "pass a knife" from top to bottom).



Let k be the point of the division. The "tip" will range from 0 to k, and the "base" from k to 4.



The volume of the tip will be:



V1 = π.∫(sqrt(x)²)dx (where the integral is from 0 to k)



and the remainder of the solid:



V2 = π.∫(sqrt(x)²)dx (where the integral is from k to 4)



Make V1 = V2 (because you want to find the value of k that makes both volumes equal), and simplify srqt(x)² to x. You will get:



π.∫ x dx = π.∫ x dx (first integral is from 0 to k, second is from k to 4)



Eliminate pi in both sides and calculate the integral on its extremes (the primitive of x is x²/2, since x²/2's derivative is x):



x²/2 | (from 0 to k) = x²/2 | (from k to 4)

k²/2 - 0 = 4²/2 - k²/2



Simplify and put k on the left side:



k² - 0 = 4² - k²

2k² = 4²

k² = 4²/2 = 8

k = √8 = 2√2



So you should "cut" the solid at the point x = 2.sqrt(2) to have both pieces with the same volume (I hope not to have screwed up the calculations, but, intuitively, it makes sense: the tip should be "taller" than the base for them to have the same volume).

Answer 3

The fundamental of pi * y^2 dx So this is going to alter into the fundamental from 0 to 3 of x*pi dx = x^2/2 * pi from 0 to 4 = 8pi it rather is even though if it fairly is revolved around the x-axis even though if it fairly is revolved around the y-axis the respond is the fundamental of pi * x^2 dy So this is going to alter into the fundamental from 0 to 2 of y^4 * pi dy = y^5/5 * pi = 32/5 * pi

Answer 4

16/3



none of the numeric restrictions have any bearing on the function, so you can just do the integral on the interval [0,4]