Find the volume formed by rotating about the y-axis the region enclosed by x=11y and y^3=x with y≥0

The curves meet at (0,0) and (11√11,√11)

So, using shells of thickness dx,
v = ∫[0,11√11] 2πrh dx
where r=x and h = ∛x - x/11
v = ∫[0,11√11] 2πx(∛x - x/11) dx = 5324/21 π√11

using washers of thickness dy,
v = ∫[0,√11] π(R^2-r^2) dy
where R = 11y and r = y^3
v = ∫[0,√11] π((11y)^2-y^6) dy = 5324/21 π√11

Oh, we're about to get a little crazy with rotating things around the y-axis! Buckle up, my friend! Now, let's dive into the world of geometry.

To find the volume formed by rotating the region enclosed by x = 11y and y^3 = x around the y-axis, we'll have to slice and dice (metaphorically speaking) our way through this problem.

First, let's find the points of intersection of these two curves. By solving the equations x = 11y and y^3 = x, we come to the conclusion that y = 1 and y = -1 are the points where they cross paths. Since we only care about the region above the x-axis (y ≥ 0), we'll stick with y = 1 for our calculations.

Now, let's set up the integral to compute the volume. We'll use the disk method because it's perfectly round and super fun!

Remember, the formula for the volume using the disk method is: V = π∫[a,b] (f(y))^2 dy.

Our bounds of integration are from y = 0 to y = 1, so our integral becomes: V = π∫[0,1] (11y)^2 dy.

Now all that's left is to integrate! After plugging in the values and performing some math magic, voila! You'll have the volume formed by rotating that freaky region around the y-axis.

P.S. If you're looking for the answer, I left that part for you to embark on this mathematical journey yourself. Trust me, it's more fun when you do it yourself and discover the answer along the way!

To find the volume formed by rotating the region enclosed by x=11y and y^3=x about the y-axis, we can use the method of cylindrical shells.

Step 1: First, we need to find the points of intersection between the two curves.

The curve x=11y intersects with the curve y^3=x when x=0:
11y=0
y=0

Let's find the other intersection point by equating the two equations:
11y=y^3
y^3-11y=0
y(y^2-11)=0

So, y=0 or y=sqrt(11) or y=-sqrt(11).

Step 2: Now, we need to determine the limits of integration for the volume integral. Since we are rotating about the y-axis, the limits of integration will be from y=0 to y=sqrt(11).

Step 3: Next, we need to find the height of each cylindrical shell. The height of each shell is determined by the difference in x-coordinates at each value of y.

The equation x=11y implies that x is equal to the radius of each cylindrical shell. So, the height of each shell is given by:
height = x = 11y

Step 4: Finally, we can set up the volume integral using the formula for cylindrical shells:
V = ∫[a,b] 2πrh dy

where r is the radius (x in this case) and h is the height.

We integrate from y=0 to y=sqrt(11), with r=11y and h=2π(11y).

V = ∫[0,sqrt(11)] 2π(11y)(2π(11y)) dy

V = ∫[0,sqrt(11)] 4π^2 * 11^2 * y^2 dy

V = 4π^2 * 11^2 * ∫[0,sqrt(11)] y^2 dy

V = 4π^2 * 11^2 * (1/3) * y^3 |[0,sqrt(11)]

V = (4/3) * π^2 * 11^2 * (sqrt(11))^3

V = (4/3) * π^2 * 11^2 * 11^(3/2)

V = (4/3) * π^2 * 11^(7/2)

Hence, the volume formed by rotating the region about the y-axis is (4/3) * π^2 * 11^(7/2).

To find the volume formed by rotating the region enclosed by x = 11y and y^3 = x about the y-axis, we can use the method of cylindrical shells.

First, let's find the points where the two curves intersect. To do this, we set them equal to each other and solve for y:

11y = y^3

Rearranging the equation, we get:

y^3 - 11y = 0

Factoring out a y from the left side:

y(y^2 - 11) = 0

Setting each factor equal to zero gives us two solutions:

y = 0 (corresponding to the y-axis)
y^2 - 11 = 0

Solving the second equation for y^2:

y^2 = 11

Taking the square root of both sides:

y = ±√11

Since we are only interested in the region where y ≥ 0, we keep y = √11 and discard y = -√11.

Now, we need to determine the limits of integration. Since we are rotating about the y-axis, the integral will be with respect to y. We integrate from y = 0 to y = √11.

The differential height of each "shell" is dy, and the radius of each shell is given by the x-value of the curve x = 11y. So, our radius is R(y) = 11y.

Now, let's express the differential volume of each shell:

dV = 2πR(y) * h * dy

Where h represents the height of each shell.

To express h in terms of y, we need to find the difference between the curves x = 11y and y^3 = x at a given value of y.

x = 11y
y^3 = x

Substituting the first equation into the second, we get:

y^3 = 11y

Dividing both sides by y:

y^2 = 11

Solving this equation, we find:

y = ±√11

Since we are rotating about the y-axis, the height of each shell is the difference between the curves evaluated at y. Therefore, h = 11y - y^3.

Now, we have everything we need to calculate the volume. We integrate the differential volume formula from y = 0 to y = √11:

V = ∫[0, √11] 2πR(y) * h * dy
= ∫[0, √11] 2π(11y) * (11y - y^3) * dy

Integrating this expression will give us the volume formed by rotating the region about the y-axis.