Applications of Integrals - Revision history

Khanh at 21:23, 28 October 2021

2021-10-28T21:23:09Z

Lila: /* Resources */

2021-10-15T20:09:03Z

Resources

Lila: /* Resources */

2021-10-15T20:07:38Z

Resources

Lila: /* Arc length */

2021-10-15T20:06:44Z

Arc length

Lila at 20:05, 15 October 2021

2021-10-15T20:05:17Z

Lila: /* Resources */

2021-10-15T20:04:04Z

Resources

Lila at 20:03, 15 October 2021

2021-10-15T20:03:20Z

Lila at 20:01, 15 October 2021

2021-10-15T20:01:33Z

Khanh: Created page with " * Average value of a function * Area between curves * Distance, velocity, acceleration * Volume * Work * Arc length * Surface area * Center of mass * Hydrostatic pressure and..."

2021-09-25T05:28:08Z

Created page with " * Average value of a function * Area between curves * Distance, velocity, acceleration * Volume * Work * Arc length * Surface area * Center of mass * Hydrostatic pressure and..."

New page

* Average value of a function
* Area between curves
* Distance, velocity, acceleration
* Volume
* Work
* Arc length
* Surface area
* Center of mass
* Hydrostatic pressure and force
* Probability

==Resources==
* [https://tutorial.math.lamar.edu/classes/calci/intappsintro.aspx Applications of Integrals - Cal I], Paul's Online Notes
* [https://tutorial.math.lamar.edu/classes/calcii/intappsintro.aspx Applications of Integrals - Cal II], Paul's Online Notes

@@ Line 122: / Line 122: @@
 * [https://en.wikibooks.org/wiki/Calculus/Volume Volume], Wikibooks: Calculus
 * [https://en.wikibooks.org/wiki/Calculus/Arc_length Arc length], Wikibooks: Calculus
 * [https://tutorial.math.lamar.edu/classes/calci/intappsintro.aspx Applications of Integrals - Cal I], Paul's Online Notes
 * [https://tutorial.math.lamar.edu/classes/calcii/intappsintro.aspx Applications of Integrals - Cal II], Paul's Online Notes

@@ Line 121: / Line 121: @@
 * [https://en.wikibooks.org/wiki/Calculus/Area Area], Wikibooks: Calculus
 * [https://en.wikibooks.org/wiki/Calculus/Volume Volume], Wikibooks: Calculus
 * [https://tutorial.math.lamar.edu/classes/calci/intappsintro.aspx Applications of Integrals - Cal I], Paul's Online Notes
 * [https://tutorial.math.lamar.edu/classes/calcii/intappsintro.aspx Applications of Integrals - Cal II], Paul's Online Notes

@@ Line 79: / Line 79: @@
 :<math>L=\int\limits_a^b \sqrt{1+\left(\tfrac{dy}{dx}\right)^2}dx</math>
-'''Proof:''' Consider <math>y_{i+1}-y_i=f(x_{i+1})-f(x_i)</math> . By the [[Calculus/Mean Value Theorem for Functions|Mean Value Theorem]] there is a point <math>z_i</math> in <math>(x_{i+1},x_i)</math> such that
+'''Proof:''' Consider <math>y_{i+1}-y_i=f(x_{i+1})-f(x_i)</math> . By the Mean Value Theorem there is a point <math>z_i</math> in <math>(x_{i+1},x_i)</math> such that
 :<math>y_{i+1}-y_i=f(x_{i+1})-f(x_i)=f'(z_i)(x_{i+1}-x_i)</math>

@@ Line 72: / Line 72: @@
 &=2\pi R^3-\frac{2\pi}{3}R^3=\frac{4\pi}{3}R^3
 \end{align}</math>
 ==Resources==

← Older revision		Revision as of 20:03, 15 October 2021
Line 15:		Line 15:

	[[File:Closed path integral defined.png\|Closed_path_integral_defined]]		[[File:Closed path integral defined.png\|Closed_path_integral_defined]]
		+
		+	==Volume==
		+	If the function <math>A(x)</math> is continuous on <math>[a,b]</math> , then the volume <math>V_S</math> of the solid <math>S</math> is given by:
		+	:{{math\|size=1.5em\|<math>V_S=\int\limits_a^b A(x)dx</math>}}
		+
		+	===Example 1: A right cylinder===
		+	[[File:Cylinder with cross section at height x.svg\|200px\|thumb\|right\|Figure 1]]
		+	Now we will calculate the volume of a right cylinder using our new ideas about how to calculate volume. Since we already know the formula for the volume of a cylinder this will give us a "sanity check" that our formulas make sense. First, we choose a dimension along which to integrate. In this case, it will greatly simplify the calculations to integrate along the height of the cylinder, so this is the direction we will choose. Thus we will call the vertical direction <math>x</math> (see [[:File:Cylinder with cross section at height x.svg\|Figure 1]]). Now we find the function, <math>A(x)</math> , which will describe the cross-sectional area of our cylinder at a height of <math>x</math> . The cross-sectional area of a cylinder is simply a circle. Now simply recall that the area of a circle is <math>\pi r^2</math> , and so <math>A(x)=\pi r^2</math> . Before performing the computation, we must choose our bounds of integration. In this case, we simply define <math>x=0</math> to be the base of the cylinder, and so we will integrate from <math>x=0</math> to <math>x=h</math> , where <math>h</math> is the height of the cylinder. Finally, we integrate:
		+	:<math>\begin{align}
		+	V_{\mathrm{cylinder}}&=\int\limits_a^b A(x)dx\\
		+	&=\int\limits_0^h\pi r^2dx\\
		+	&=\pi r^2\int\limits_0^hdx\\
		+	&=\pi r^2x\bigg\|_{x=0}^h\\
		+	&=\pi r^2(h-0)\\
		+	&=\pi r^2h\end{align}</math>
		+
		+	This is exactly the familiar formula for the volume of a cylinder.
		+
		+	===Example 2: A right circular cone===
		+	[[File:Plane intersecting cone 2.png\|200px\|thumb\|right\|Figure 2: The cross-section of a right circular cone by a plane perpendicular to the axis of the cone is a circle.]]
		+
		+	For our next example we will look at an example where the cross sectional area is not constant. Consider a right circular cone. Once again the cross sections are simply circles. But now the radius varies from the base of the cone to the tip. Once again we choose <math>x</math> to be the vertical direction, with the base at <math>x=0</math> and the tip at <math>x=h</math> , and we will let <math>R</math> denote the radius of the base. While we know the cross sections are just circles we cannot calculate the area of the cross sections unless we find some way to determine the radius of the circle at height <math>x</math> .
		+
		+	[[File:Similar triangles for cone.svg\|200px\|thumb\|right\|Figure 3: Cross-section of the right circular cone by a plane perpendicular to the base and passing through the tip.]]
		+
		+	Luckily in this case it is possible to use some of what we know from geometry. We can imagine cutting the cone perpendicular to the base through some diameter of the circle all the way to the tip of the cone. If we then look at the flat side we just created, we will see simply a triangle, whose geometry we understand well. The right triangle from the tip to the base at height <math>x</math> is similar to the right triangle from the tip to the base at height <math>h</math> . This tells us that <math>\frac{r}{h-x}=\frac{R}{h}</math> . So that we see that the radius of the circle at height <math>x</math> is <math>r(x)=\frac{R}{h}(h-x)</math> . Now using the familiar formula for the area of a circle we see that <math>A(x)=\pi\frac{R^2}{h^2}(h-x)^2</math> .
		+
		+	Now we are ready to integrate.
		+
		+	:<math>\begin{align}
		+	V_{\mathrm{cone}}&=\int\limits_a^b A(x)dx\\
		+	&=\int\limits_0^h \pi\frac{R^2}{h^2}(h-x)^2dx\\
		+	&=\pi\frac{R^2}{h^2}\int\limits_0^h(h-x)^2dx
		+	\end{align}</math>
		+
		+	By u-substitution we may let <math>u=h-x</math> , then <math>du=-dx</math> and our integral becomes
		+
		+	:<math>\begin{align}
		+	&&=\pi\frac{R^2}{h^2}\left(-\int\limits_h^0 u^2du\right)\\
		+	&&=\pi\frac{R^2}{h^2}\left(-\frac{u^3}{3}\bigg\|_h^0\right)\\
		+	&&=\pi\frac{R^2}{h^2}\left(-0+\frac{h^3}{3}\right)\\
		+	&&=\frac{\pi}{3}R^2h
		+	\end{align}</math>
		+
		+	===Example 3: A sphere===
		+	[[File:Sphere with cross section.svg\|200px\|thumb\|right\|Figure 4: Determining the radius of the cross-section of the sphere at a distance <math>\|x\|</math> from the sphere's center.]]
		+	In a similar fashion, we can use our definition to prove the well known formula for the volume of a sphere. First, we must find our cross-sectional area function, <math>A(x)</math> . Consider a sphere of radius <math>R</math> which is centered at the origin in <math>\R^3</math> . If we again integrate vertically then <math>x</math> will vary from <math>-R</math> to <math>R</math> . In order to find the area of a particular cross section it helps to draw a right triangle whose points lie at the center of the sphere, the center of the circular cross section, and at a point along the circumference of the cross section. As shown in the diagram the side lengths of this triangle will be <math>R</math> , <math>\|x\|</math> , and <math>r</math> . Where <math>r</math> is the radius of the circular cross section. Then by the Pythagorean theorem <math>r=\sqrt{R^2-\|x\|^2}</math> and find that <math>A(x)=\pi(R^2-\|x\|^2)</math> . It is slightly helpful to notice that <math>\|x\|^2=x^2</math> so we do not need to keep the absolute value.
		+
		+	So we have that
		+	:<math>\begin{align}
		+	V_{\mathrm{sphere}}&=\int\limits_a^b A(x)dx\\
		+	&=\int\limits_{-R}^R\pi(R^2-x^2)dx\\
		+	&=\pi\int\limits_{-R}^R R^2dx-\pi\int\limits_{-R}^R x^2dx\\
		+	&=\pi R^2x\Bigg\|_{-R}^R-\pi\frac{x^3}{3}\Bigg\|_{-R}^R\\
		+	&=\pi R^2(R-(-R))-\pi\left(\frac{R^3}{3}-\frac{(-R)^3}{3}\right)\\
		+	&=2\pi R^3-\frac{2\pi}{3}R^3=\frac{4\pi}{3}R^3
		+	\end{align}</math>

	==Resources==		==Resources==