Math / Geometry

Question14. [0.5/1 Points] DETAILS MY NOTES SCALC9 5.3.064. PREVIOUS ANSWERS ASK YOUR TEACHER PRACTICE ANOTHER
Suppose you make napkin rings by drilling holes with different diameters through two wooden balls (which also have different diameters). You discover that both napkin rings have the same height ih, as shown in the figure. (Let $a=7$ .) (a) Which ring has more wood in it? The napkin ring created by drilling a hole with the larger radius has more wood in it. The amount of wood in each napkin ring is the same regardless of the size of the sphere used. The napkin ring created by drilling a hole with the smaller radius has more wood in it. (b) Use cylindrical shells to compute the volume $V$ of a napkin ring of height $7 h$ created by driling a hole with radius $r$ through the center of a sphere of radius $R$ and express the answer in terms of $h$ . $v=$ $\square$
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Studdy Solution

STEP 1

What is this asking? We're comparing the volume of napkin rings made from spheres with different sizes and holes with different radii, and then we need to calculate the volume of a napkin ring with a height of $7h$ . Watch out! Don't assume the bigger sphere means more wood in the ring!
The size of the hole plays a crucial role.
Also, remember we're dealing with volumes, not just areas.

STEP 2

1. Compare the rings
2. Set up the integral
3. Solve the integral
4. Substitute and simplify
5. Adjust for the new height

STEP 3

Surprisingly, both napkin rings have the same amount of wood!
This is a classic result.
It seems counterintuitive, but the math will show us why!

STEP 4

To find the volume, we'll use the method of cylindrical shells.
Imagine slicing the napkin ring into super thin, nested cylinders.

STEP 5

The volume of a cylindrical shell is $2 \cdot \pi \cdot x \cdot y \cdot dx$ , where $x$ is the radius of the shell, $y$ is the height of the shell, and $dx$ is the tiny thickness.

STEP 6

In our case, the radius of the shell is just $x$ .
The height of the shell, $y$ , is $2\sqrt{R^2 - x^2}$ (twice the height of the right triangle formed by $R$ , $x$ , and half of $y$ ).

STEP 7

Our limits of integration are from the radius of the hole, $r$ , to the radius of the sphere, $R$ .
So, the integral for the volume $V$ is: $V = \int_r^R 2 \cdot \pi \cdot x \cdot 2\sqrt{R^2 - x^2} \, dx = 4\pi \int_r^R x\sqrt{R^2 - x^2} \, dx$

STEP 8

U-substitution time!
Let $u = R^2 - x^2$ .
Then $du = -2x \, dx$ , so $x \, dx = -\frac{1}{2} du$ .

STEP 9

Our integral becomes: $V = 4\pi \int_{R^2 - r^2}^0 \sqrt{u} \left( -\frac{1}{2} \right) du = -2\pi \int_{R^2 - r^2}^0 u^{1/2} \, du$

STEP 10

Now we can integrate: $V = -2\pi \left[ \frac{2}{3} u^{3/2} \right]_{R^2 - r^2}^0 = -\frac{4\pi}{3} \left[ 0 - (R^2 - r^2)^{3/2} \right] = \frac{4\pi}{3} (R^2 - r^2)^{3/2}$

STEP 11

Notice that $(R^2 - r^2)$ is the square of half the height, $(\frac{h}{2})^2 = \frac{h^2}{4}$ .

STEP 12

Substituting this back into our volume equation: $V = \frac{4\pi}{3} \left( \frac{h^2}{4} \right)^{3/2} = \frac{4\pi}{3} \cdot \frac{h^3}{8} = \frac{\pi h^3}{6}$

STEP 13

The problem asks for the volume with a height of $7h$ .
Let's call this new volume $V_{7h}$ .

STEP 14

We just replace $h$ with $7h$ in our formula: $V_{7h} = \frac{\pi (7h)^3}{6} = \frac{343\pi h^3}{6}$

STEP 15

The volume of a napkin ring with height $h$ is $\frac{\pi h^3}{6}$ , regardless of the sphere and hole sizes.
With a height of $7h$ , the volume is $\frac{343\pi h^3}{6}$ .

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Studdy Solution

STEP 1

STEP 2

1. Compare the rings
2. Set up the integral
3. Solve the integral
4. Substitute and simplify
5. Adjust for the new height

STEP 3

Surprisingly, both napkin rings have the same amount of wood!
This is a classic result.
It seems counterintuitive, but the math will show us why!

STEP 4

To find the volume, we'll use the method of cylindrical shells.
Imagine slicing the napkin ring into super thin, nested cylinders.

STEP 5

The volume of a cylindrical shell is $2 \cdot \pi \cdot x \cdot y \cdot dx$ , where $x$ is the radius of the shell, $y$ is the height of the shell, and $dx$ is the tiny thickness.

STEP 6

In our case, the radius of the shell is just $x$ .
The height of the shell, $y$ , is $2\sqrt{R^2 - x^2}$ (twice the height of the right triangle formed by $R$ , $x$ , and half of $y$ ).

STEP 7

Our limits of integration are from the radius of the hole, $r$ , to the radius of the sphere, $R$ .
So, the integral for the volume $V$ is: $V = \int_r^R 2 \cdot \pi \cdot x \cdot 2\sqrt{R^2 - x^2} \, dx = 4\pi \int_r^R x\sqrt{R^2 - x^2} \, dx$

STEP 8

U-substitution time!
Let $u = R^2 - x^2$ .
Then $du = -2x \, dx$ , so $x \, dx = -\frac{1}{2} du$ .

STEP 9

Our integral becomes: $V = 4\pi \int_{R^2 - r^2}^0 \sqrt{u} \left( -\frac{1}{2} \right) du = -2\pi \int_{R^2 - r^2}^0 u^{1/2} \, du$

STEP 10

Now we can integrate: $V = -2\pi \left[ \frac{2}{3} u^{3/2} \right]_{R^2 - r^2}^0 = -\frac{4\pi}{3} \left[ 0 - (R^2 - r^2)^{3/2} \right] = \frac{4\pi}{3} (R^2 - r^2)^{3/2}$

STEP 11

Notice that $(R^2 - r^2)$ is the square of half the height, $(\frac{h}{2})^2 = \frac{h^2}{4}$ .

STEP 12

Substituting this back into our volume equation: $V = \frac{4\pi}{3} \left( \frac{h^2}{4} \right)^{3/2} = \frac{4\pi}{3} \cdot \frac{h^3}{8} = \frac{\pi h^3}{6}$

STEP 13

The problem asks for the volume with a height of $7h$ .
Let's call this new volume $V_{7h}$ .

STEP 14

We just replace $h$ with $7h$ in our formula: $V_{7h} = \frac{\pi (7h)^3}{6} = \frac{343\pi h^3}{6}$

STEP 15

The volume of a napkin ring with height $h$ is $\frac{\pi h^3}{6}$ , regardless of the sphere and hole sizes.
With a height of $7h$ , the volume is $\frac{343\pi h^3}{6}$ .

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Studdy Solution

STEP 1

STEP 2

1. Compare the rings2. Set up the integral3. Solve the integral4. Substitute and simplify5. Adjust for the new height

STEP 3

Surprisingly, both napkin rings have the same amount of wood!This is a classic result.It seems counterintuitive, but the math will show us why!

STEP 4

To find the volume, we'll use the **method of cylindrical shells**.Imagine slicing the napkin ring into super thin, nested cylinders.

STEP 5

The volume of a cylindrical shell is 2⋅π⋅x⋅y⋅dx2 \cdot \pi \cdot x \cdot y \cdot dx2⋅π⋅x⋅y⋅dx, where xxx is the radius of the shell, yyy is the height of the shell, and dxdxdx is the tiny thickness.

STEP 6

In our case, the radius of the shell is just xxx.The height of the shell, yyy, is 2R2−x22\sqrt{R^2 - x^2}2R2−x2​ (twice the height of the right triangle formed by RRR, xxx, and half of yyy).

STEP 7

STEP 8

**U-substitution** time!Let u=R2−x2u = R^2 - x^2u=R2−x2.Then du=−2x dxdu = -2x \, dxdu=−2xdx, so x dx=−12dux \, dx = -\frac{1}{2} duxdx=−21​du.

STEP 9

Our integral becomes: V=4π∫R2−r20u(−12)du=−2π∫R2−r20u1/2 duV = 4\pi \int_{R^2 - r^2}^0 \sqrt{u} \left( -\frac{1}{2} \right) du = -2\pi \int_{R^2 - r^2}^0 u^{1/2} \, duV=4π∫R2−r20​u​(−21​)du=−2π∫R2−r20​u1/2du

STEP 10

STEP 11

Notice that (R2−r2)(R^2 - r^2)(R2−r2) is the square of half the height, (h2)2=h24(\frac{h}{2})^2 = \frac{h^2}{4}(2h​)2=4h2​.

STEP 12

Substituting this back into our volume equation: V=4π3(h24)3/2=4π3⋅h38=πh36V = \frac{4\pi}{3} \left( \frac{h^2}{4} \right)^{3/2} = \frac{4\pi}{3} \cdot \frac{h^3}{8} = \frac{\pi h^3}{6}V=34π​(4h2​)3/2=34π​⋅8h3​=6πh3​

STEP 13

The problem asks for the volume with a height of 7h7h7h.Let's call this new volume V7hV_{7h}V7h​.

STEP 14

We just replace hhh with 7h7h7h in our formula: V7h=π(7h)36=343πh36V_{7h} = \frac{\pi (7h)^3}{6} = \frac{343\pi h^3}{6}V7h​=6π(7h)3​=6343πh3​

STEP 15

The volume of a napkin ring with height hhh is πh36\frac{\pi h^3}{6}6πh3​, regardless of the sphere and hole sizes.With a height of 7h7h7h, the volume is 343πh36\frac{343\pi h^3}{6}6343πh3​.

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1. Compare the rings
2. Set up the integral
3. Solve the integral
4. Substitute and simplify
5. Adjust for the new height

Surprisingly, both napkin rings have the same amount of wood!
This is a classic result.
It seems counterintuitive, but the math will show us why!

To find the volume, we'll use the method of cylindrical shells.
Imagine slicing the napkin ring into super thin, nested cylinders.

The volume of a cylindrical shell is $2 \cdot \pi \cdot x \cdot y \cdot dx$ , where $x$ is the radius of the shell, $y$ is the height of the shell, and $dx$ is the tiny thickness.

In our case, the radius of the shell is just $x$ .
The height of the shell, $y$ , is $2\sqrt{R^2 - x^2}$ (twice the height of the right triangle formed by $R$ , $x$ , and half of $y$ ).

U-substitution time!
Let $u = R^2 - x^2$ .
Then $du = -2x \, dx$ , so $x \, dx = -\frac{1}{2} du$ .

Our integral becomes: $V = 4\pi \int_{R^2 - r^2}^0 \sqrt{u} \left( -\frac{1}{2} \right) du = -2\pi \int_{R^2 - r^2}^0 u^{1/2} \, du$

Notice that $(R^2 - r^2)$ is the square of half the height, $(\frac{h}{2})^2 = \frac{h^2}{4}$ .

Substituting this back into our volume equation: $V = \frac{4\pi}{3} \left( \frac{h^2}{4} \right)^{3/2} = \frac{4\pi}{3} \cdot \frac{h^3}{8} = \frac{\pi h^3}{6}$

The problem asks for the volume with a height of $7h$ .
Let's call this new volume $V_{7h}$ .

We just replace $h$ with $7h$ in our formula: $V_{7h} = \frac{\pi (7h)^3}{6} = \frac{343\pi h^3}{6}$

The volume of a napkin ring with height $h$ is $\frac{\pi h^3}{6}$ , regardless of the sphere and hole sizes.
With a height of $7h$ , the volume is $\frac{343\pi h^3}{6}$ .