Worksheets/Archimdes Dont Disturb My Circles - Revision history

Admin: /* Part 3: Surface Area of a Sphere */

2016-05-09T07:23:52Z

Part 3: Surface Area of a Sphere

← Older revision		Revision as of 07:23, 9 May 2016
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	Part 3: How to obtain the surface area of a sphere from the volume of a sphere (via differentiation)		Part 3: How to obtain the surface area of a sphere from the volume of a sphere (via differentiation)

	~~Start~~ with volume of a sphere:		To find surface area, we can note that surface area is equal to the rate of change of volume as we shrink the radius:

			<math>
			SA = \dfrac{dv}{dr}
			</math>

			Now, let's start with the equation for the volume of a sphere:

	<math>		<math>
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	</math>		</math>

	~~Now differentiate~~ both sides. We'll find the change in volume that corresponds to a shell of thickness dr:		Differentiate both sides. We'll find the change in volume that corresponds to a shell of thickness dr:

	<math>		<math>

Admin: /* Questions */

2016-05-09T07:22:53Z

Questions

← Older revision		Revision as of 07:22, 9 May 2016
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	2) What is the integral that relates the circumference of a circle to the area of a circle?		2) What is the integral that relates the circumference of a circle to the area of a circle?

	3) Mr. Blue is applying the disk method to compute the volume of a sphere with radius R. Mr. Blue takes a slice of the sphere to form a disk with a width of <math>\Delta x</math> and a cross-sectional area of <math>\pi R^2</math>. Mr. Blue then computes the area of the sphere using the integral <math>\int_{0}^{R}{\pi R^2 dx}, yielding the volume of a sphere of <math>V = \pi R^3</math>. What conceptual error did Mr. Blue make?		3) Mr. Blue is applying the disk method to compute the volume of a sphere with radius R. Mr. Blue takes a slice of the sphere to form a disk with a width of <math>\Delta x</math> and a cross-sectional area of <math>\pi R^2</math>. Mr. Blue then computes the area of the sphere using the integral <math>\int_{0}^{R}{\pi R^2 dx}</math>, yielding the volume of a sphere of <math>V = \pi R^3</math>. What conceptual error did Mr. Blue make?

	4) Write the integral equation for a volume using the most general expression for the area A(x) as a function of location.		4) Write the integral equation for a volume using the most general expression for the area A(x) as a function of location.

Admin: /* Questions */

2016-05-09T07:22:41Z

Questions

← Older revision		Revision as of 07:22, 9 May 2016
Line 119:		Line 119:
	2) What is the integral that relates the circumference of a circle to the area of a circle?		2) What is the integral that relates the circumference of a circle to the area of a circle?

	3) Mr. Blue is applying the disk method to compute the volume of a sphere with radius R. Mr. Blue takes a slice of the sphere to form a disk with a width of <math>\Delta x</math> and a cross-sectional area of <math>\pi R^2</math>. Mr. Blue then computes the area of the sphere using the integral <math>\int_{0}^{R} \pi R^2 dx}, yielding the volume of a sphere of <math>V = \pi R^3</math>. What conceptual error did Mr. Blue make?		3) Mr. Blue is applying the disk method to compute the volume of a sphere with radius R. Mr. Blue takes a slice of the sphere to form a disk with a width of <math>\Delta x</math> and a cross-sectional area of <math>\pi R^2</math>. Mr. Blue then computes the area of the sphere using the integral <math>\int_{0}^{R}{\pi R^2 dx}, yielding the volume of a sphere of <math>V = \pi R^3</math>. What conceptual error did Mr. Blue make?

	4) Write the integral equation for a volume using the most general expression for the area A(x) as a function of location.		4) Write the integral equation for a volume using the most general expression for the area A(x) as a function of location.

Admin: /* Questions */

2016-05-09T07:22:29Z

Questions

← Older revision		Revision as of 07:22, 9 May 2016
Line 119:		Line 119:
	2) What is the integral that relates the circumference of a circle to the area of a circle?		2) What is the integral that relates the circumference of a circle to the area of a circle?

	3) Mr. Blue is applying the disk method to compute the volume of a sphere with radius R. Mr. Blue takes a slice of the sphere to form a disk with a width of ~~%DELTA~~ x and a cross-sectional area of %pi R^2. Mr. Blue then computes the area of the sphere using the integral ~~int from~~{0} to{R} {%pi R^2 dx}, ~~thus~~ yielding the volume of a sphere of %pi R^3. What conceptual error did Mr. Blue make?		3) Mr. Blue is applying the disk method to compute the volume of a sphere with radius R. Mr. Blue takes a slice of the sphere to form a disk with a width of <math>\Delta x</math> and a cross-sectional area of <math>\pi R^2</math>. Mr. Blue then computes the area of the sphere using the integral <math>\int_{0}^{R} \pi R^2 dx}, yielding the volume of a sphere of <math>V = \pi R^3</math>. What conceptual error did Mr. Blue make?

	4) Write the integral equation for a volume using the most general expression for the area A(x) as a function of location.		4) Write the integral equation for a volume using the most general expression for the area A(x) as a function of location.

Admin: /* Part 3: Surface Area of a Sphere */

2016-05-09T07:21:26Z

Part 3: Surface Area of a Sphere

← Older revision		Revision as of 07:21, 9 May 2016
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	<math>		<math>
	dV = \frac{4}{3} \pi 3 r^2 dr		dV = \frac{4}{3} \pi 3 r^2 dr = 4 \pi r^2 dr
	~~</math>~~

	~~or,~~

	~~<math>~~
	dV = 4 \pi r^2 dr
	</math>		</math>

Admin: /* Part 3: Surface Area of a Sphere */

2016-05-09T07:21:12Z

Part 3: Surface Area of a Sphere

← Older revision		Revision as of 07:21, 9 May 2016
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	Part 3: How to obtain the surface area of a sphere from the volume of a sphere (via differentiation)		Part 3: How to obtain the surface area of a sphere from the volume of a sphere (via differentiation)

			Start with volume of a sphere:

			<math>
			V = \frac{4}{3} \pi r^3
			</math>

			Now differentiate both sides. We'll find the change in volume that corresponds to a shell of thickness dr:

			<math>
			dV = \frac{4}{3} \pi 3 r^2 dr
			</math>

			or,

			<math>
			dV = 4 \pi r^2 dr
			</math>

			or,

			<math>
			\frac{dV}{dr} = 4 \pi r^2
			</math>

	=Questions=		=Questions=

Admin: /* Part 2: Volume of a Sphere */

2016-05-09T07:19:19Z

Part 2: Volume of a Sphere

← Older revision		Revision as of 07:19, 9 May 2016
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	<math>V = \int_{-r}^{r} \pi \sqrt{r^2 - x^2}^2 dx</math>		<math>V = \int_{-r}^{r} \pi \sqrt{r^2 - x^2}^2 dx</math>

	<math>V = \pi \left( r^2 x - \~~dfrac~~{x^3}{3} right)</math>		<math>V = \pi \left( r^2 x - \frac{x^3}{3} \right) \bigg\|_{-r}^{r}</math>

	<math>V = 2 \pi \left( r^3 - \~~dfrac~~{r^3}{3} \right) = \dfrac{4}{3} \pi r^3 </math>		<math>V = 2 \pi \left( r^3 - \frac{r^3}{3} \right) = \dfrac{4}{3} \pi r^3 </math>

	This yields the formula for the volume of a sphere:		This yields the formula for the volume of a sphere:

Admin: /* Part 2: Volume of a Sphere */

2016-05-09T07:17:08Z

Part 2: Volume of a Sphere

← Older revision		Revision as of 07:17, 9 May 2016
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	<math>V = \int_{-r}^{r} \pi \sqrt{r^2 - x^2}^2 dx</math>		<math>V = \int_{-r}^{r} \pi \sqrt{r^2 - x^2}^2 dx</math>

	<math>V = \pi \left( r^2 x - \dfrac{x^3}{3~~} \bigg\|_{-r}^{r~~} right)</math>		<math>V = \pi \left( r^2 x - \dfrac{x^3}{3} right)</math>

	<math>V = 2 \pi \left( r^3 - \dfrac{r^3}{3} \right) = \dfrac{4}{3} \pi r^3 </math>		<math>V = 2 \pi \left( r^3 - \dfrac{r^3}{3} \right) = \dfrac{4}{3} \pi r^3 </math>

Admin: /* Part 2: Volume of a Sphere */

2016-05-09T07:16:47Z

Part 2: Volume of a Sphere

← Older revision		Revision as of 07:16, 9 May 2016
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	Limits of integration for x are -r to r, which make the disk method:		Limits of integration for x are -r to r, which make the disk method:

	<math>V = int_{-r}^{r} \pi \sqrt{r^2 - x^2}^2 dx</math>		<math>V = \int_{-r}^{r} \pi \sqrt{r^2 - x^2}^2 dx</math>

	<math>V = \pi \left( r^2 x - \dfrac{x^3}{3} \bigg\|_{-r}^{r} right) = 2 \pi (r^3 - \dfrac{r^3}{3}) = \dfrac{4}{3} \pi r^3 </math>		<math>V = \pi \left( r^2 x - \dfrac{x^3}{3} \bigg\|_{-r}^{r} right)</math>

			<math>V = 2 \pi \left( r^3 - \dfrac{r^3}{3} \right) = \dfrac{4}{3} \pi r^3 </math>

	This yields the formula for the volume of a sphere:		This yields the formula for the volume of a sphere:

Admin: /* Part 2: Volume of a Sphere */

2016-05-09T07:16:12Z

Part 2: Volume of a Sphere

← Older revision		Revision as of 07:16, 9 May 2016
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	The circle equation is given by <math>x^2 + y^2 = r^2</math> or <math>y = \pm \sqrt{r^2 - x^2}</math>.		The circle equation is given by <math>x^2 + y^2 = r^2</math> or <math>y = \pm \sqrt{r^2 - x^2}</math>.

			Limits of integration for x are -r to r, which make the disk method:

			<math>V = int_{-r}^{r} \pi \sqrt{r^2 - x^2}^2 dx</math>

			<math>V = \pi \left( r^2 x - \dfrac{x^3}{3} \bigg\|_{-r}^{r} right) = 2 \pi (r^3 - \dfrac{r^3}{3}) = \dfrac{4}{3} \pi r^3 </math>

			This yields the formula for the volume of a sphere:

			<math>V = \dfrac{4}{3} \pi r^3</math>

	===Part 3: Surface Area of a Sphere===		===Part 3: Surface Area of a Sphere===