Volume of revolution around the y-axis

[Kqwert]

Homework Statement

Hello,

a bowl is created when rotating the function

$$f(x) = \begin{cases} 0, & 0 \leq x \lt 6 \\ (12/\pi)arcsin(x-6), & 6\leq x \leq 7 \end{cases}$$

around the y-axis. Find the height (h) and the volume (V) of the bowl.

Homework Equations

The Attempt at a Solution

So, I graphed the problem and found the height to be 6 (maximum y-value from the arcsin function). Apparently this is wrong. Why?
For the volume, I considered the integrals
$$\int_0^6 2\pi(6-0)x \, dx$$ and $$2\pi \int_6^7 (2/\pi)arcsin(x-6)x \, dx$$

and then adding them together. Where did I go wrong?

 

[Mark44]

Homework Statement

Hello,

a bowl is created when rotating the function

$$f(x) = \begin{cases} 0, & 0 \leq x \lt 6 \\ (12/\pi)arcsin(x-6), & 6\leq x \leq 7 \end{cases}$$

around the y-axis. Find the height (h) and the volume (V) of the bowl.

Homework Equations

The Attempt at a Solution

So, I graphed the problem and found the height to be 6 (maximum y-value from the arcsin function). Apparently this is wrong. Why?

Why do you think the height is wrong? The max. value from the arcsin function is 1, and the max. value from 12/pi arcsin(x - 6) is attained when x = 7.

For the volume, I considered the integrals
$$\int_0^6 2\pi(6-0)x \, dx$$ and $$2\pi \int_6^7 (2/\pi)arcsin(x-6)x \, dx$$

and then adding them together.

The first integral doesn't make sense, nor does adding them together.

Where did I go wrong?

You don't have the radius right -- it should be x + 6.

Apparently you are using shells to get the volume. Start with a reasonably accurate sketch of the region that will be revolved. This will be a roughly triangular region. A typical area element is a thin vertical strip going from the x-axis to the arcsine curve. When this strip is revolved around the y-axis, it forms a thin shell. What is the volume of this shell volume element?

 

[Kqwert]

For both of the integrals or only the last one?

 

[Mark44]

For both of the integrals or only the last one?

I've edited my earlier post, so take a look at it again. Your first integral doesn't make any sense.

 

[Kqwert]

The area bounded by the red and green lines are our area, right? (see thumbnail) If so, then from x = 6 to x = 7 the volume should be
$$\int_6^7 2\pi(6-(12/\pi)arcsin(x-6))(6+x) \, dx$$

Correct..?

 

[Mark44]

The area bounded by the red and green lines are our area, right? (see thumbnail) If so, then from x = 6 to x = 7 the volume should be
$$\int_6^7 2\pi(6-(12/\pi)arcsin(x-6))(6+x) \, dx$$

Correct..?

Looks good to me...

 

[Kqwert]

Looks good to me...

Thanks! but what about the "center" of the bowl? i.e. 0 < x < 6. I guess I can solve this without integration, but how do I solve it w integration?

 

[Mark44]

Thanks! but what about the "center" of the bowl? i.e. 0 < x < 6. I guess I can solve this without integration, but how do I solve it w integration?

As I understand the problem, the bottom of the "bowl" is the plane formed from rotation the x-axis around the y-axis. Using the shell method, the region of integration is bounded by the lines x = 6, x = 7, and the curve. The shells are generated from thin strips between x = 6 and x = 7, so those are the limits of integration.

If this isn't the region of integration, but is instead the region bounded by the lines x = 6, y = 6, and the curve, then revolving that region around the y-axis gives a whole different solid of revolution, with a different volume. Is there a picture in your textbook of what the solid should look like?

 

[Kqwert]

No, it´s just given as I described in the first post.

 

[Mark44]

a bowl is created when rotating the function

The problem statement as posted isn't clear to me. I've been assuming that the sort of triangular region bounded by the x-axis, the line x = 7, and the curve is revolved around the y-axis. This would make a figure whose outer surface is a cylinder and whose inner surface is concave.

The integral you set up made me think that's what the solid is.

If the inner surface looks the same as the outer surface, then your first integral in post #1 looks fine, but the second integral doesn't. You are apparently using thin vertical strips, but between x = 6 and x = 7, your strips run from the x-axis up to the arcsine curve, which is wrong. They should instead run from the curve up to the line y = 6.

 

[Kqwert]

But wouldn't the area between the line y = 6 and the arcsin curve, between x = 6 and x = 7, be the integral in post #5?

 

[Ray Vickson]

Thanks! but what about the "center" of the bowl? i.e. 0 < x < 6. I guess I can solve this without integration, but how do I solve it w integration?

The center is a standard cylinder. You probably learned how to compute its volume years before you studied calculus----it is just high-school geometry. In any case, just about every calculus textbook ever written will have sections in which such simple volumes are computed by integration.

 

[Kqwert]

Yeah, still get the wrong answer though, not sure if I am computing the correct volume.

 

[Mark44]

But wouldn't the area between the line y = 6 and the arcsin curve, between x = 6 and x = 7, be the integral in post #5?

Yes, and the total volume would be that plus the volume of the cylinder in the middle.

 

[LCKurtz]

I'm a bit late to this thread, but surely wouldn't the natural way to set up this problem be using disks and a dy integral:
$$V = \int_0^6 \pi \left ( 6+\sin(\frac{\pi }{12}y)\right)^2~dy$$

 

[Mark44]

I'm a bit late to this thread, but surely wouldn't the natural way to set up this problem be using disks and a dy integral

Same thought occurred to me, as well.

 

[Kqwert]

In the curriculum we have only been through the disk method for rotation around the x-axis and cylindrical shells for rotation around the y-axis.

 

[Ray Vickson]

In the curriculum we have only been through the disk method for rotation around the x-axis and cylindrical shells for rotation around the y-axis.

So, what is stopping you from using either one on this problem?

You have seen both methods, and there is no good reason for limiting yourself to only one kind for any given problem.

 

[SammyS]

The area bounded by the red and green lines are our area, right? (see thumbnail) If so, then from x = 6 to x = 7 the volume should be
$$\int_6^7 2\pi(6-(12/\pi)arcsin(x-6))(6+x) \, dx$$

Correct..?

This is even later to the thread.

Here you have the correct height for a shell,

$\left(6-\frac{12}{\pi}\arcsin(x-6)\right)$​

but the radius for a shell is $\ x\$ as you had originally.

This gives the following for the volume generated by the portion of the function lying between x=6 and x=7.
$\displaystyle{\int_6^7} \left(2 \pi \left(6-\frac{12}{\pi}\arcsin(x-6)\right)(x)\right) \, dx$
.

 

[Kqwert]

Thanks!

but what about the inner portion of the bowl? i.e. from 0 < x < 6. Is it pi*6^3..?

 

[SammyS]

Thanks!

but what about the inner portion of the bowl? i.e. from 0 < x < 6. Is it pi*6^3..?

From Mark44 (post #10):

...

If the inner surface looks the same as the outer surface, then your first integral in post #1 looks fine, ...

So, the integral $\displaystyle\ \int_0^6 2\pi(6-0)x \, dx\$ gives the volume swept out by (height −ƒ(x)) for 0 ≤ x ≤ 6, where the height is 6 units. That is $\ 6^3\pi\$ as you asked.