Volume of Solids?

Question

I need to find the volume formed by the region enclosed by:

x=3y and y^3 = x with y being greater than or equal to 0 about the y-axis.

I know I have to take the difference of the integral of 3/x from x^(1/3)

but after that, I have no idea what to do.

Similarly, I need to find the volume of the solid formed by rotating the region inside the first quadrant enclosed by:

y=x^4
y=64x


They're very frustrating because I have an idea of how to do them but, I don't really know how to apply the integrations.
AH!

Answer 1

Use the washer method.

x = 3y
x = y³

Set the two equations equal to find the limits of integration.

3y = y³
y³ - 3y = 0
y(y² - 3) = 0
y(y - √3)(y + √3) = 0
y = 0, √3
Since y ≥ 0.

Integrate from y = 0 to √3.

Volume = ∫π(R² - r²)dy = π∫[(3y)² - (y³)²]dy

= π∫[9y² - y^6]dy

= π[3y³ - y^7/7] | [Evaluated from 0 to √3]

= π[3^(5/2) - (1/7)3^(7/2)] = 9π√3[1 - 3/7]

= 9π√3[4/7] = 36π√3 / 7

Answer 2

Sketching a picture would help a lot.

For the first problem, we're only dealing with y being nonnegative. However, this also means that x must be nonnegative (because a positive number cubed will always yield a positive number). Thus, we're working in the first quadrant only.

Next, we need to find the intersection points.

Set y^3=3y

y^3-3y=y(y^2-3)=0

and we get y=0 and y=sqrt(3) (we throw out the negative answer).

Thus the intersecting points are (0,0) and (3sqrt(3), sqrt(3)).

Graphing this tells us that x=3y is to the right of y^3=x (we're not concerned with which one is "higher" because we're revolving about a vertical line. Thus the integrand would be:

pi[(3y)^2-(y^3)^2]=pi(9y^2-y^6)

and the limits would be 0 to sqrt(3).

Answer 3

First, you need to sketch the region enclosed by these curves.
You also need to get the points of intersection of the two.
3y = y³ the intersection occur at y=0 & y = √3.

Note also that x=y³ is to the left of x=3y.

We can set-up the integral for the volume of the solid from the region revolved about the y-axis using the washer method. (This means the partitions are horizontal, we use dy.)

V = π ∫[0.. √3] (3y)²-(y³)² dy [Edit: sorry, wrong position earlier]

you can finish this now...



for the next question, use dx for the differential because the functions are expressed in terms of x.
again get x-values of the intersection.
determine which function is above the other.
(what is the axis of revolution?)


d:

Answer 4

First of all, find the intersection points of the curves. For the first one, x = 3y = y^3 => y = 0 or ±√3; since y ≥ 0 that means we have the interval [0, √3] for y. In this interval 3y ≥ y^3 ≥ 0. We're rotating around the y-axis, so we have
V = π∫x^2 dy
= π∫(0 to √3) ((3y)^2 - (y^3)^2) dy
= π (9y^3 / 3 - y^7 / 7)[0 to √3]
= π(9√3 - 27√3 / 7)
= 36π√3 / 7.

For the second one, you don't say what axis you're rotating it around; I shall assume the x-axis. Again, find the intersection points: (0, 0) and (4, 256).
So we have V = π∫(0 to 4) ((64x)^2 - (x^4)^2) dx
= π(4096x^3/3 - x^9/9) [0 to 4]
= π(262144/3 - 262144/9)
= 524288π/9.

Answer 5

The triangle is a suitable triangle, a^2+b^2=c^2. this means that, once you placed the triangle over the x axis, it will be defined via a single function that's a today line that is going interior the path of the factors (0,0) and (15,20). This function is f(x) = 4/3x. Now you calculate the quantity of the forged of revolution with the formulation V = ? ? f(x)^2 dx. Which, for this reason is ? * the fundamental from 0 to fifteen of sixteen/9x^2 dx, = ?16/27x^3 evaluated at 15 and nil = ?16/27*15^3 = 2000?. thank you very lots!

Answer 6

Imagine they are f(x) & g(x). you must write:
h(x)=f(x)-g(x)
then h(x)=0 find the x! they must be at least 2 answers for x
for example x^4-64x=0 then x=0 or x=4
then u find somthing I din't know what it is in english. but it's the integratin. u must do it on the h(x) which is f(x)-g(x) or g(x)-h(x). really there is no diffrence to use aby of them but use those x u found as domains
wish u got your answer. if not, please contact me!
titish.lovely