--- ## The basics - Using integration we can **find the volume of any mathematically defined shape.** For our purposes this means: if a shape's **cross sectional area can be described by a function,** we can find its area through integration - To find the volume of an object using integration we conceptually divide that object into a number of individual **slices**. We then find the **cross sectional area of these "slices" and multiply that by their thickness** to get the volume of each slice. The summation of these volumes should be equal to the **approximate volume of the shape**. $\newline$ $\text{Using a finite number of slices:}$ $\color{grey}\small\textit{let }x_i^*=\textit{ x position of a slice }\space\space\space\textit{let }\Delta x_i=\textit{ distance between slices: }\Delta x_i=\frac{\textit{number of slices}}{b-a}$ $\newline$ $\text{Volume of the }n^{\text{th}}\text{ slice }=V(n)\approx A(x_i^*)\cdot\Delta x_i$ $\text{Thus the total volume }= \color{LightCyan}V\approx\sum_{i=1}^n A(x_i^*)\cdot\Delta x_i$ $\newline$ - To get the **exact volume of a 3 dimensional solid** we use the same process, except ==**instead of evaluating a finite number of "slices" we use integration to evaluate the volume of an *infinite number of infinitely small slices.***== - To visualize this process imagine moving a flat plane through your object and collecting the cross sectional area "projected" onto that plane at every moment. The plane will move along the axis **specified in the integrand from the lower bound to the upper bound**. - If the integrand **is in terms of x the plane will "move" along the x axis** if the integrand is in terms of y **it will "move" along the y axis**. - If your integrand is in terms of **x then the upper and lower bounds must be x values**. If it is is in terms of y then the bounds must be **y values**. - The summation equation above adds up the volumes of a discrete number of slices. In order to use this "scanner" method we must have an **infinite** number of slices which is why we use integration: $\newline$ $\large V=\lim_{\Delta x\rightarrow 0}\left(\sum_{i=1}^nA(x_i^*)\cdot\Delta x_i\right)\longrightarrow\color{Bisque}\int_a^b A(x)dx$ $\newline$ >[!caution] >- **The integrand and the bounds must both refer to values along the same axis**. >- Therefore, if the integrand **is in terms of y the bounds *($a$ and $b$)* are also in terms of y**. If the integrand is in terms of x, the bounds are also in terms of x. >[!example] >##### *Volume of a sphere* >- As a 2D plane, moving along the x axis travels through a sphere, the sphere will project a circle of slowly increasing, and then decreasing radii. >- Since we know the **radius of the sphere and the x position of this "plane" we can use the Pythagorean theorem to derive the radius of the *projected circle* in terms of x** *(and the sphere's radius)*. > >$\newline$ >$\biggl(x^2+y^2=r^2\biggl)\longrightarrow\biggl(y=\sqrt{r^2-x^2}\biggl)$ >$\therefore A(x)=\pi\cdot y^2\longrightarrow\text{\color{grey}\small substitute}\longrightarrow A(x)=\pi(r^2-x^2)$ >$\newline$ >- Now that we have an equation for the cross sectional area of the circle in terms of known variables and x we can move on to integration. However, before that we need to figure out one more thing, the bounds. Since this sphere is centered at $(0,0,0)$, we know that the leftmost edge of the sphere is at $x=-r$ and the rightmost side of the sphere is at $x=r$. Therefore our bounds are $-r$ and $r$. > >$\newline$ >$V=\int_{-r}^r A(x)\space dx\longrightarrow\int_{-r}^r \pi(r^2-x^2)dx=2\pi\int_{0}^r(r^2-x^2)dx$ >$=2\pi\Biggl[r^2x-\frac{x^3}{3}\Biggl]_0^r\longrightarrow 2\pi\Biggl(r^3-\frac{r^3}{3}\Biggl)=\color{Bisque}\frac{4}{3}\pi\cdot r^3$ >$\newline$ >--- > >![[Volume and integrationddd.png|850]] --- ## Wolfram alpha >[!cite] >--- > ><iframe src="https://www.wolframalpha.com/input?i=solid+of+revolution&assumption=%22ClashPrefs%22+-%3E+%7B%22MathWorld%22%2C+%22SolidofRevolution%22%7D" width="850" height = "800" style="position: relative; left: 0px; top: 0px" ></iframe> --- ## Basic solids - A "basic solid" is a 3 dimensional shape that is created by **giving a 2 dimensional area thickness**. - To find the volume of a *"basic"* solid **we imagine a scanner moving through the solid and recording the cross sectional area of the shape at every point.** If the area enclosed by the original function is extended out into the z axis uniformly, the 2D shape "projected" onto the "scanner" will be a rectangle. The height of this **rectangle will be equal to the function's output at that x value and the width of this rectangle will be a constant value: the thickness of the solid.**. - By integrating the equation for the cross sectional area of the shape between two bounds, we can find the **volume of the solid formed between those bounds**. $A(x)=d\cdot f(x)\space\space{\color{grey}\small(\text{where }d\text{ is the depth of the solid})}$ $\large\therefore V(x)=\int A(x)dx=d\int f(x)dx$ >[!quote] >##### *Basic solid visualization* >![[Volume and integration example2.png|850]] --- ## Solids of revolution - A solid of revolution is a 3 dimensional shape that is created using the outline **of a function that has been rotated around an axis**. For example, if the function $y=-x+5$ is "rotated" around the *y axis* it will define the shape of an upright cone. - This process of creating shapes is useful **because it allows us to describe complex 3 dimensional shapes using a single 2 dimensional function**, like $y=mx+b$. - By definition, *solids of revolution* are **rotationally symmetric** along their axis of revolution. In other words, if you were to split the object in half using a plane that pivots about it's axis of revolution, the volumes created on either side would be identical no mater how the plane is rotated. --- >[!tip] >![[solids or rotation.png|850]] --- ## *The disk method* - The cross sectional area of solids **that touch their axis of revolution at all points** *(like the one shown above)* is equal to a circle of radius $f(x)$. Taking the integral of this cross sectional area will give us the volume of its corresponding shape. - Imagine scanning the solid of revolution shown above using an upright plane that's moving along the x axis from left to right. Its fairly easy to see that the shape projected onto that plane will be a circle at all points and the radius of this circle will be directly proportional to $f(x)$, the solids defining function. - When using the disk method the function/integrand **must be in terms of the same variable as the axis of rotation** *(x or y).* - This is known as the "disk method" and it only applies to solids of revolution that **touch their axis of rotation** at **all points on the bound**. $\newline$ $\text{Cross sectional area using "disk" method: } \pi r^2\rightarrow A(x)=\pi\Big(f(x)\Big)^2$ $\small\color{grey}\textit{where }f(x)\textit{ is function describe the solid of rotation}$ $\text{Volume of each disk: }V(x)=A(x)\Delta x\longrightarrow\color{LightCyan}V=\int A(x)dx$ $\newline$ >[!quote] >- Find the volume of the solid of rotation created when the function $\sqrt{x}$ is rotated around the x axis between the bounds $[0,1]$. >--- >$A(x)=\pi\Big(f(x)\Big)^2\longrightarrow A(x)=\pi\Big(\sqrt{x}\Big)^2=\pi x$ > $\therefore V=\int_0^1 A(x)dx=\int_0^1\pi x\space dx\rightarrow\text{\small\color{grey}(integrate)}\rightarrow\Biggl[\pi\frac{x^2}{2}\Biggl]_0^1=\color{Bisque}\frac{\pi}{2}$ > $\large\color{Bisque}\textit{The volume is }V=\frac{\pi}{2}$ > > ![[74f1aa81df528275f24c8887051c7fac.gif|850]] --- ## The "Washer method" - The disk method outlined above **only works if the solid of revolution *touches* its axis of rotation at all points on the bound.** To find the volume of a solid of revolution that **doesn't touch its rotational axis at *all* points** we have to utilize a different technique. This section outlines one alternative technique that is known colloquially, as the *"washer method"*. - Imagine a plane traveling through a solid of revolution that **does not touch the axis on which it was rotated**. Instead of "projecting" a circle onto this plane the solid will project **a ring**. - If the axis of rotation is x, the **integrand must be in terms of y** for this method to work *(and the other way around)*. Therefore, we can say that this method **requires the the axis of revolution and the integrand to be in terms of opposite axis.** - To calculate the area of a ring we **subtract the area of a circle that extends to its inner edge from the area of a circle that extends to its outer edge.** The radius of the inner circle is *usually* equal to the solid's corresponding function $f(x)$ and the radius of the outer circle is *usually* just a constant value, equal to one of the bounds. - If the axis of revolution is not $x=0$ or $y=0$ you must add or subtract some constant value from the tube's radius to correct for the axis of revolution's offset. $\newline$ $\text{Cross sectional area using "washer" method: }A(x)=\pi\Big(f(x)\Big)^2-\pi\Big(g(x)\Big)^2$ $\small\color{grey}\textit{where }g(x)\textit{ is the function giving the area of the inner circle and }f(x)\textit{ is the outer circle}$ $\therefore V=\int A(x)dx=\pi\int\big( f(x)\big)^2dx-\pi\int\big(g(x)\big)^2dx$ $\newline$ >[!quote] >- What is the volume of the solid of revolution created by rotating the function $y=x$ around the y axis on the bounds X: $[0,3]$ >--- >$\text{Area of outter circle: }f(x)=\pi(6)^2=36\pi\space\space\textit{\small\color{grey} (the area is constant)}$ >$\text{Area of inner circle: }g(x)=\pi y^2$ >$V=\int_0^3 36-\int_0^3\pi y^2dy\longrightarrow(36*3)-28.27=\color{Bisque}79.73$ > >![[e0684ecaef307d3abba5d03aa2356a28.gif|850]] --- ## The tube/shell method - As the name suggests this method involves splitting a solid of revolution up into an **infinite** number of tubes. Since the thickness of these tubes must be infinitely small they exist only as 2 dimensional areas, equal to the product of **their circumference and height**. By taking the integral of the "tubes" we can find the volume of their corresponding solid. - This method is similar to the *"washer method"* since it can be used to find the volume of SOR's that **don't touch their axis of revolution on all points between the specified bounds**. - If the axis of rotation is x, the **integrand must also be in terms of x** for this method to work *(the same applies to y)*. Therefore, we can say that this method **requires the the axis of revolution and the integrand to be in terms of the same axis.** - The **height of each tube is equal to the function's output at its location $x$**. The circumference of the tube is equal to the circumference of a circle with a radius equal to the tube's distance from the axis of revolution. - If the axis of revolution is not $x=0$ or $y=0$ you must add or subtract some constant value from the tube's radius to correct for the axis of revolution's offset. $\newline$ $\text{Cross sectional area using the "tube" method: }A(x)=(2\pi x)\cdot f(x)$ $\large\therefore V=\int A(x)\space dx=2\pi\int xf(x)\space dx$ $\newline$ >[!quote] >##### *Visual of the tube/shell method* ![[Volume and integrationsdfsdf3342.png|850]] --- ## Solids of revolution using two functions - Using integration we can find the volume of any solid created by **rotating the area between two functions about an axis**. Using two functions instead of one to describe the shape of a 2 dimensional area we can make 3 dimensional solids of revolution with more complex geometries. - The area contained **between two functions on the 2 dimensional plane will almost never touch the axis on which it is being rotated**. As a result, we only use either the, "washer" or "tube" method to find the volume of these objects. - The "larger" function is usually denoted as $f(x)$. If the integrand is in terms of x this is the function whose output is larger *in the positive y direction (up)*. If the integrand is in terms of y this is the function whose output is further in the *positive x direction (right)*. - Its important to recall that this changes every time the two lines intersect. Therefore, a separate integral must be used for **each pair of intersections**. >[!Danger] >- If you are using the washer method, its important to first make sure that neither of the two functions **overlap themselves horizontally on the specified bounds**. A function overlaps *horizontally* if it is simultaneously equal to **the same y value more than once**. If this happens and you insist on using the washer method you will have to evaluate the volume of the overlapping section *separately* *(make sure to adjust the bounds accordingly)* since the area equations will be different. >- Similarly, if you are using the *tube* method its important to ensure that neither of the two function's **overlap themselves *vertically*, that is be equal to the same x value more than once.** Again, if you have to use tubes you'll need to evaluate separate integrals, one for the non overlapping section, and one for the overlapping section. > >##### *An example of a horizontally overlapping function (blue)* >![[Volume and integrationoverlapingstuff.png|850]] ##### *Using the washer method* - This method should be used when the **radius of the inner and outer circles each correspond to a separate function.** This happens primarily when the enclosed area is relatively parallel to its axis of revolution. - The cross sectional area of a ring is equal to **the area of a circle that extends to its outer edge minus the area of a circle that extends to its inner edge**. The radius of each circle must be given in the form of an equation. This equation must output the **distance between the axis of rotation and its corresponding edge *(either inner or outer)*. - Like everything else in this section, you can integrate the equation this equation to find the *volume* of the overall solid. - If the axis of revolution is not $x=0$ or $y=0$ you must add or subtract some constant value from the tube's radius to correct for the axis of revolution's offset. $A(x)=\pi\Big(f(x)\Big)^2-\pi\Big(g(x)\Big)^2$ $\small\color{grey}f(x)\textit{ is the larger function and }g(x)\text{ is the smaller function}$ $\large\therefore V=\int A(x)dx=\pi\int \Big(f(x)\Big)^2dx\space -\pi\int\Big(g(x)\Big)^2$ $\newline$ >[!quote] >- Find the volume of a solid of revolution with an inner radius of $1+y$ and an outer radius of $1+\sqrt{y}$ on the bounds [0,1] when it is rotated about $x=-1$ >--- >$A(x)=\pi(\textit{outer radius})^2-\pi(\textit{inner radius})^2\longrightarrow A(x)=\pi\Big(1+\sqrt{y}\Big)^2-\pi\Big(1+y\Big)^2$ >$V=\int_0^1A(y)dy\longrightarrow\pi\int_0^1\biggl[(1+\sqrt{y})^2-(1+y)^2\biggl]dy$ >$\therefore\pi\Biggl[\frac{4y^{\frac{3}{2}}}{3}-\frac{y^2}{2}-\frac{y^3}{3}\Biggl]_0^1\rightarrow\text{\color{grey}\small integrate}\rightarrow\color{Bisque} A=\frac{\pi}{2}$ >![[7d8e65264040385d053ce21ba04cb4ef.gif|850]] ##### *Using the tube method* - The tube method should be used when the **two functions that enclose an area each *separately* correspond to the bottom and top of a tube that's centered around the axis of revolution.** This usually happens when the area enclosed by the two functions is *relatively* perpendicular to the axis of revolution. - Recall that the tube method requires us to find both the circumference and height of a perfectly thin cylinder at any point $x$. The radius of the solid is still just **the distance from the axis of revolution to the tube's position, $x$.** The height of the tube is equal to **the height of the larger function $f(x)$ minus the height of the smaller function.** - Like everything else in this section, you can integrate the equation this equation to find the *volume* of the overall solid. - If the axis of revolution is not $x=0$ or $y=0$ you must add or subtract some constant value from the tube's radius to correct for the axis of revolution's offset. $A(x)=(2\pi x)\big(f(x)-g(x)\big)$ $\small\color{grey}\textit{where }f(x)\textit{ is the larger function in either the positive x or y direction}$ $\large\therefore V=\int A(x)dx=2\pi\int x\Big(f(x)-g(x)\Big)dx$ >[!quote] >- Use the tube method to find the volume generated by rotating the region bounded by $y=2x^2$ and $y=12x-4x^2$ is rotated about the _y_-axis. >--- >##### *Step 1 - Find the bounds* >- We can tell from the functions that $y=12x-4x^2$ is the "upper function" since it is an upside down parabola which encloses the bottom part of the upright parabola that corresponds with $y=2x^2$. Now I set the equations equal and find the bounds. > >$\begin{aligned}2x^2&=12x-4x^2\\\\{\color{grey}(+4x^2)}\space2x^2&=12x-\cancel{4x^2}\space{\color{grey}(+4x^2)}\\\frac{\cancel{6}x^2}{\color{grey}6}&=\frac{12\cancel{x}}{\color{grey}x}\\x&=2\\\therefore x&=0\end{aligned}$ >##### *Step 2 - Find the equation for area:* >- If I were to solve this problem using the *washer method* I would have to split the problem up into two integrals, therefore I am using the tube method. This works because there is no place on the bounds where the functions overlap themselves vertically, *along the y axis*. > >$\text{Circumference }\rightarrow c=2\pi x$ >$\text{Height }\rightarrow f(x)-g(x)=\Big(12x-4x^2\Big)-\Big(2x^2\Big)=h$ >$\small\color{grey}f(x) \text{ is the larger function and }g(x)\text{ is the smaller function}$ >$\large\therefore A(x)=h\cdot c=(2\pi x)\Big[(12x-4x^2)-(2x^2)\Big]$ > >##### *Step 3 - Integration* >- Now that we have an equation for **the area of a 2D "tube" at any value of x** we can find the solid's volume by integrating the area equation. I usually split my integral into 2 parts using the **addition/subtraction property** since it keeps my work more organized. >$V=\left(2\pi\int_0^2 12x-4x^2\space dx\right)\space -\left(2\pi\int_0^2 2x^2\space dx\right)$ >$\newline$ >$\text{Work for the first integral:}$ >$\left(2\pi\int_0^2 12x-4x^2\space dx\right)\longrightarrow 2\pi\left[6x^2-\frac{4}{3}x^3+C\right]_0^2$ >$\therefore V_1=2\pi\biggl(6(2)^2-\frac{4}{3}(2)^3\biggl)-2\pi\big(0\big)=\color{LightCyan}\frac{80\pi}{3}$ >$\newline$ >$\text{Work for second integral:}$ >$\left(2\pi\int_0^2 2x^2\space dx\right)\longrightarrow 2\pi\biggl[\frac{2}{3}x^3\biggl]_0^2\longrightarrow 2\pi\biggl(\frac{2}{3}(2)^3\biggl)-2\pi\big(0\big)=\color{LightCyan}\frac{32\pi}{3}$ >$\newline$ >$\large\color{Bisque}\text{Thus the volume of the solid is: }\left(\frac{80\pi}{3}-\frac{32\pi}{3}\right)=\frac{48\pi}{3}=16\pi$ > --- #mainpage