I would give this video a million likes if I could. I am having to code this and for the life of me could not understand it, but now I do. thank you so much.
why are we not infegrating pie? or is pie integrated is pie itself?(this is a stupid question so please don't laugh at me) ( 8:37 ), I think that the integration of pie needs to be xπ because pie is just a number. (please answer this question 🥺)
You would do that if it were an addition of pie and the (r^2-x^2). However, here it is pi TIMES (r^2-x^2) , so you antiderive the (r^2-x^2) and leave the constant (pi) as it is. I think this is one of the antiderivative rules. Hope it helps!
Pi is a constant, so you can pull it out in front of the integral. Just as if you were given k*x^2 to integrate, you can multiply k by the integral of x^2.
Dont confuse urself. Look, he is integrating x. (Dx) Since x has a power of 0 on r2, when integrating, it becomes 0+1 = 1 The 1 divides by r2 and x to the power 1 is 1 Thus u get r2x/1 = r2x Just take r2 as a common number like 3 When u integrate 3, u get 3x
Terry Shannon The volume of the sphere is the sum of the cylinders you get rotating the little rectangles from the integral of the function around the *x* axis. Let me explain it: first of all, the function we are interested in is *y = sqrt(r² - x²)* (I'll explain why at the end), so the integral of that function would be the sum of little rectangles with width *dx* and height *y* (and remember that *y = sqrt(r² - x²)* ). So the area of the rectangles are *y·dx* , and the sum of all of them (the integral) is the area under the function. But we want a volume of the sphere, and how did we get the sphere? We rotated the.function (the semicircle) around the *x* axis, so the volume would be equal to that area rotated around the *x* axis. And remember the area is a bunch of rectangles, so the volume will be equal to the sum of those rectangles rotated around the *x* axis. And if you rotate a rectangle you get a cylinder. So the volume is equal to the sum of all the cylinders. The volume of a cylinder is equal to pi multiplied by its height and multiplied by its radius squared. Since we are rotating around the *x* axis the radius (the moving distance) is equal to *y* , and the height is a little portion ( infinitesimal) of the *x* axis: I mean *dx* . So the volume of a cylinder at any point will be *pi·dx·y²* , and if you rearrange: *pi·y²·dx* . Remember *y = sqrt(r² -x²)* , so finally we get *pi·(r²-x²)·dx* . But we want the volume of the whole sphere, so we want the sum of all those cylinders, and that's the integral from *-r* to *r* of *pi·(r²-x²)·dx* . Solve it and you get the formula! Now I'll answer your other question. I think you're confusing the parabola function and the circle function. They look alike, but they're not the same function. The equation of the parabola is *y = x²* and that means that for any *x* value, *y* is that *x* squared. If you plot the pair of values *(x,y)* you get the parabola. But the circle is something different. Its equation is *y = sqrt(r²-x²)* , but actually the most known form of it is *y² + x² = r²* . Let's take a look at it. First of all, *r* is just a number, not a variable. Its called *r* because it's the radius of the circle. So think of *r* as a fixed distance from the point *(0,0)* . It can be 1, it can be 5... But it's not a variable, it's a number you choose before plotting the circle function. Then we know that a circle with center at *(0,0)* and radius *r* is the set of points with a distance *r* from the center. And how can we write that? Well, take any point of the circle, then draw a line from it to the center (the radius), and draw another line from it to the *x* axis (perpendicularly). It's a triangle! The base is *x* , the height is *y* , and the hypotenuse is *r* . How can we relate those things? With the pythagorean theorem! Then you see that *y² + x² = r²* is the way to plot all the pair of values *(x,y)* (all the points) with a distance *r* (a number) from the center ( *(0,0)* ). And that's why *y = sqrt(r²-x²)* . It's not the same as a parabola.
Finally, if you're wondering why the equation only takes care of the up-semicircle and not the whole circle; it is by convention. Mathematicians agreed that, by definition, plotting a function *y* in terms of *x* could only have one solution for *y* given any *x* value. Nevertheless, *x* could have more solutions: when *y = x²* if you let *y = 4* , then could be *x = 2* , but also *x = -2* . But if *y² = x* and then *y = sqrt(x)* , for any value *x* there's only one solution (only if we are considering it a function). I know that for *x = 4* you could get *y = 2* and *y = -2*, but because of the definition of a function, we get rid of the negative value. And if you think about it, it makes sense, because if you plot *y = sqrt(x)* with the negative values for *x* included, then you cannot calculate the area under the function, because the positive area above the *x* axis would cancel out with the negative area below the *x* axis and you'll always get *0* . So in order to be possible to calculate an area (with the integral), we have to ignore the negative part of *y = sqrt(x)* . And the same happens with the circle: if you calculate the area of the circle as a function, till you'll get *0* because the up-area would cancel out with the down-area. So we have to get rid of the negative part of the circle, and we only consider the semicircle as the function from the equation *y² + x² = r²* . That's all.
In simple words, apply the pythagorus theorem Keep a right angled triangle And ur work is done Take base as x Take perpendicular as y And take hypotenuse as r
I just found this channel and I gotta say, this is some of the most entertaining math I've seen in a long time.
SIR PLEASE DONT STOP MAKING VIDEOS. THESE ARE REALLY HELPFUL
4:50 the upper boundary is supposed to be 9 not 2, for anyone confused
Why doesn't this have like a million views??
Jesus christ, what a teacher!
right?!
WOW! Such a succint explanation for the volume of a sphere! Blew my mind. Love it!
How I love math. Watching your videos at 5am. If my math classes was made simple when I was studying in high school.
yes better than rote memorizing steps
okay, WOW i wish you were my math teacher. time to transfer schools...
Inspiring as always. Keep up the good work sir!!
I would give this video a million likes if I could. I am having to code this and for the life of me could not understand it, but now I do. thank you so much.
How's your brain doing? 😅 Love it mr. Woo.
Just great! Thank you.
this is amazing. thank you!!
This is awesome
The limits in terms of y should be 9 and 0. This is still amazing🥰🥰
if i had to describe the latter half of this video in 3 words it would be "cathartic sphere mindblowing"
Very nice explanation, thank you.
why are we not infegrating pie? or is pie integrated is pie itself?(this is a stupid question so please don't laugh at me) ( 8:37 ), I think that the integration of pie needs to be xπ because pie is just a number. (please answer this question 🥺)
You would do that if it were an addition of pie and the (r^2-x^2). However, here it is pi TIMES (r^2-x^2) , so you antiderive the (r^2-x^2) and leave the constant (pi) as it is. I think this is one of the antiderivative rules. Hope it helps!
Sorry if im late but thats probably because pie is not part of the original function so thats why we dont integrate it ( ur question was nice tho)
Pi is a constant, so you can pull it out in front of the integral. Just as if you were given k*x^2 to integrate, you can multiply k by the integral of x^2.
spongooo!
Amazing
Isn't it supposed to be the square root of y??? x^2=y... x=sqrt(y)......
ay^2=x^3 ko x=0 to x= 4a rotated about y axis then value of surface area ...
Sir please make video on thi
Isn't it suppose to be y^3, not y^2?
The bounds for you fist solid of revolution problem is 0-9 not 0-2.
your* first*
who said it was 0-2?
4:49
I didnt understand the step when he integrated r^2 = r^2 x
Dont confuse urself. Look, he is integrating x. (Dx)
Since x has a power of 0 on r2, when integrating, it becomes 0+1 = 1
The 1 divides by r2 and x to the power 1 is 1
Thus u get r2x/1 = r2x
Just take r2 as a common number like 3
When u integrate 3, u get 3x
revelation !
Why is the volume of the sphere the integral of pi y^2 dx? For the parabola y = x^2, but here y = sqrt(r^2 - x^2)?
Terry Shannon The volume of the sphere is the sum of the cylinders you get rotating the little rectangles from the integral of the function around the *x* axis.
Let me explain it: first of all, the function we are interested in is *y = sqrt(r² - x²)* (I'll explain why at the end), so the integral of that function would be the sum of little rectangles with width *dx* and height *y* (and remember that *y = sqrt(r² - x²)* ). So the area of the rectangles are *y·dx* , and the sum of all of them (the integral) is the area under the function.
But we want a volume of the sphere, and how did we get the sphere? We rotated the.function (the semicircle) around the *x* axis, so the volume would be equal to that area rotated around the *x* axis. And remember the area is a bunch of rectangles, so the volume will be equal to the sum of those rectangles rotated around the *x* axis. And if you rotate a rectangle you get a cylinder. So the volume is equal to the sum of all the cylinders.
The volume of a cylinder is equal to pi multiplied by its height and multiplied by its radius squared. Since we are rotating around the *x* axis the radius (the moving distance) is equal to *y* , and the height is a little portion ( infinitesimal) of the *x* axis: I mean *dx* . So the volume of a cylinder at any point will be *pi·dx·y²* , and if you rearrange: *pi·y²·dx* . Remember *y = sqrt(r² -x²)* , so finally we get *pi·(r²-x²)·dx* . But we want the volume of the whole sphere, so we want the sum of all those cylinders, and that's the integral from *-r* to *r* of *pi·(r²-x²)·dx* . Solve it and you get the formula!
Now I'll answer your other question. I think you're confusing the parabola function and the circle function. They look alike, but they're not the same function. The equation of the parabola is *y = x²* and that means that for any *x* value, *y* is that *x* squared. If you plot the pair of values *(x,y)* you get the parabola.
But the circle is something different. Its equation is *y = sqrt(r²-x²)* , but actually the most known form of it is *y² + x² = r²* . Let's take a look at it. First of all, *r* is just a number, not a variable. Its called *r* because it's the radius of the circle. So think of *r* as a fixed distance from the point *(0,0)* . It can be 1, it can be 5... But it's not a variable, it's a number you choose before plotting the circle function. Then we know that a circle with center at *(0,0)* and radius *r* is the set of points with a distance *r* from the center. And how can we write that? Well, take any point of the circle, then draw a line from it to the center (the radius), and draw another line from it to the *x* axis (perpendicularly). It's a triangle! The base is *x* , the height is *y* , and the hypotenuse is *r* . How can we relate those things? With the pythagorean theorem! Then you see that *y² + x² = r²* is the way to plot all the pair of values *(x,y)* (all the points) with a distance *r* (a number) from the center ( *(0,0)* ). And that's why *y = sqrt(r²-x²)* . It's not the same as a parabola.
Finally, if you're wondering why the equation only takes care of the up-semicircle and not the whole circle; it is by convention. Mathematicians agreed that, by definition, plotting a function *y* in terms of *x* could only have one solution for *y* given any *x* value. Nevertheless, *x* could have more solutions: when *y = x²* if you let *y = 4* , then could be *x = 2* , but also *x = -2* .
But if *y² = x* and then *y = sqrt(x)* , for any value *x* there's only one solution (only if we are considering it a function). I know that for *x = 4* you could get *y = 2* and *y = -2*, but because of the definition of a function, we get rid of the negative value. And if you think about it, it makes sense, because if you plot *y = sqrt(x)* with the negative values for *x* included, then you cannot calculate the area under the function, because the positive area above the *x* axis would cancel out with the negative area below the *x* axis and you'll always get *0* . So in order to be possible to calculate an area (with the integral), we have to ignore the negative part of *y = sqrt(x)* . And the same happens with the circle: if you calculate the area of the circle as a function, till you'll get *0* because the up-area would cancel out with the down-area. So we have to get rid of the negative part of the circle, and we only consider the semicircle as the function from the equation *y² + x² = r²* . That's all.
In simple words, apply the pythagorus theorem
Keep a right angled triangle
And ur work is done
Take base as x
Take perpendicular as y
And take hypotenuse as r
I didn't like maths much but now.....