- Washer method rotating around horizontal line (not x-axis), part 1
- Washer method rotating around horizontal line (not x-axis), part 2
- Washer method rotating around vertical line (not y-axis), part 1
- Washer method rotating around vertical line (not y-axis), part 2
- Washer method: revolving around other axes
Doing some hairy algebra and arithmetic to evaluate the definite integral from the last video. Created by Sal Khan.
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- How would you find the area of an equation/equations that was rotated around a line that wasn't horizontal or vertical?(5 votes)
- I might sound silly but why cant we just sink the figure into water and measure how much water fell of?(2 votes)
- Also, the water method (if applicable) should be performed at constant temperature to avoid change in water density. If temperature changes, the whole method will fail. However, using calculus (the beauty of mathematics) we can find volume of shapes in matter of minutes, anywhere.(6 votes)
- If you were to revolve the functions about y=-4, would you find the same volume?(3 votes)
- It would NOT be for two reasons. since the general formula is (top curve- bottom curve) you would subtract a negative 4 instead of a positive one.Also, the curve causing the gap between the axis of rotation would be different. in the video, the curve causing the gap, or empty space would be y=x, and the washer's height would be dictated by the other function. in your question, it would be the other way around. the integral set up would be the integral (from 0 to 3) of [(x+4)^2-(x^2-2x+4)].(3 votes)
- When you raise x to the fourth at a little before6:00and the result is 81, wouldn't that negative sign become positive? ie Yielding positive 81 in lieu of negative 81??(3 votes)
- It would become positive if we had (-x)^4, because then we'd be multiplying -x by itself four times, But in this video we have -x^4 instead. The way to evaluate that expression is to multiply x by itself four times and then applying the minus sign. So when x = 3 we get -(3*3*3*3) which is -81.(2 votes)
- Is the same if you put the 4 at the end of each equation instead that at the beginning?
for example: x^2-2x-4 instead of 4-x^2+2x(1 vote)
- Sal did put the polynomial into standard form before he squared it.
His order was -x² + 2x + 4
Yes, the result is exactly the same when you square x²- 2x- 4 and compare with the square of 4 - x²+ 2x and the square of -x² + 2x + 4. This is also true if you change the signs of 4 - x before squaring.
For many purposes, you can switch the signs of the coefficients as long as you switch all of them. You can solve for x, for example after switching all of the signs. You will get the same answer when you square the polynomials after switching the signs:
(-x² +2x +4)² is x⁴ - 4x³ - 4x² + 16x + 16
(x²- 2x- 4)² is x⁴ - 4x³ - 4x² + 16x + 16
likewise (4 - x)² = (x - 4)² = x² - 8x + 16
But when finding integrals, you should keep the signs the way they were.
∫ (-x² + 2x + 4) dx
= ∫ -x² dx + ∫ 2x dx + ∫ 4 dx
= - ⅓ x³ + x² + 4x
If you evaluate the integral between 0 and 4, the result is -64/3 + 32 = 32/3
∫ (x² - 2x - 4) dx
= ∫ x² dx - ∫ 2x dx - ∫ 4 dx
= ⅓ x³ - x² - 4x
If you evaluate the integral between 0 and 4, the result is 64/3 - 32= -32/3, the exact opposite of the correct answer.
Also, make sure you do not try to graph after switching the signs: the shape will be flipped.(4 votes)
- when subtracting the 4-x^2+2x..... shouldn't it be 4-(x^2+2x) and the negative sign is distributed? Can you please explain this- thank you!(2 votes)
- No because when you remove the negative from (-x^2) and (2x), you would get (x^2) and (-2x), so you would get 4-x^2+2x=4-(x^2-2x). You are distributing the negative sign to both values in the now separated polynomial by making the second value negative.(1 vote)
- What is being done when substracting the function from 4 isn't basically working with the figure on the y=0 axis? Please correct me if wrong, thank you.(1 vote)
- For the most part, just ignore the x-axis. The act of subtracting either function from 4 gives you the actual magnitude of the radius for both the inner and outer circle.
Take y = x for example as the radius of the inner circle. At x = 1, the y value is 1. The axis of rotation is y = 4 so the y-value is 4. The radius of the inner circle is the difference between these two values all the way around. 4 - 1 = 3 at that value. At x = 2, the y-value of the function is 2 and so the radius is 4 - 2 = 2.
The same thing applies to y = x^2 - 2x. Subtracting the value of this function from 4 gives you the magnitude of the radius of the outer circle. At x = 1, the y-value is -1. Since we don't care about negative values in this application because we want positive (real) volume of solids, subtracting from 4 gives us the magnitude of this distance. So 4 - -1 = 5. It's a distance of 4 from y = 4 to y = 0 plus the additional distance from y = 0 to y = -1. That total distance gives us a radius of 5.
If you want to think about the concept in general, just forget about the y and x axes and think of y = 4 as the only axis. Subtracting from that starting point will give you the difference from the starting point to the function. That difference is what we care about since that represents the radius that we are rotating around the origin ( y = 4).(3 votes)
- Could you do U-substitution or something instead of multiplying the polynomial?(0 votes)
- I got confused on the part where you began to add them. I distributed my negative sign before adding and got the opposite answer. Was I not supposed to do that. Instead of 4-x, I got x-4 and then did the multiplication.(1 vote)
- For this example why did we not have to account for the area under the x axis formed by the curve y= x^2 -2x in the definite integral. From my understanding that area is negative and we have to take the absolute value of that interval after finding the rest of the area. Does it not matter for the washer/ disk method?(1 vote)
Where we left off in the last video, we'd actually set up our definite integral to figure out the volume of this figure. So now we just have to evaluate it. And really the hardest part is going to be simplifying that and that right over there. So let's get to it. So what is this thing squared? Well, looks like we're going to have to do a little bit of polynomial multiplication here. So I'll go into that same color I had. So we're going to have 4 minus x squared plus 2x times 4 minus-- actually, let me write that in the order of the terms, or the degree of the terms. It's negative x squared plus 2x plus 4-- I just switched the order of these things-- times negative x squared plus 2x plus 4. So we're just going to multiply these two things. I shouldn't even write a multiplication symbol, it looks too much like an x. So 4 times 4 is 16. 4 times 2x is 8x. 4 times negative x squared is negative 4x squared. 2x times 4 is 8x. 2x times 2x is 4x squared. And then 2x times negative x squared is negative 2x to the third power. And now we just have to multiply negative x times all this. So negative-- or negative x squared. Negative x squared times 4 is negative 4x squared. Negative x squared times 2x is negative 2x to the third power. And then negative x squared times negative x squared is positive x to the fourth. So it's going to be positive x to the fourth. And now we just have to add up all of these terms. And we get, let's see, we get x to the fourth-- add these two-- minus 4x to the third. And then this cancels with this, but we still have that, so minus 4x squared. You add these two right over here, you get plus 16x. And then you have plus 16. So that's this expanded out. And now if we want to-- 4 minus x times 4 minus x. So if we just have 4 minus x times 4 minus x, we could actually do it this way as well. But that's just going to be 4 times 4, which is 16, plus 4 times negative x, which is negative 4x, negative x times 4, another negative 4x, and then negative x times negative x is equal to plus x squared. So if we were to swap the order, we get x squared minus 8x plus 16. But we're going to subtract this. So if you have the negative sign out there, we're going to subtract this business. So let's just do it right over here since we already have it set up. So we have this minus this. We're going to subtract this, or we could add the negative of it. So we'll put negative x squared plus 8x minus 16. So I'm just going to add the negative of this. And we get-- and I'll do this in a new color-- let's see, we get x to the fourth minus 4x to the third power minus 5x squared plus 24x, and then these cancel out. So that's what we are left with. And so that's the inside of our integral. So we're going to take the integral of this thing, just so I don't have to keep rewriting this thing, from 0 until, if I remember correctly, 3. Yup, from 0 to 3, of this, dx. And then we had a pi out front here. So I'll just take that out of the integral. So times pi. Well, now we just to take the antiderivative. And this is going to be equal to pi times antiderivative of x to the fourth is x to the fifth over 5. Antiderivative of 4x to the third is actually x to the fourth. So this is going to be minus x to the fourth-- you can verify that. Derivative is 4x, and then you decrement the exponent, 4x to the third. So that works out. And then the antiderivative of this is negative 5/3 x to the third. Just incremented the exponent and divided by that. And then you have plus 24x squared over 2, or 12x squared. And we're going to evaluate that. I like to match colors for my opening and closing parentheses. We're going to evaluate that-- actually, let me do it as brackets. We're going to evaluate that from 0 to 3. So this is going to be equal to pi times-- let's evaluate all this business at 3. So we're going to get 3 to the fifth power. So let's see, 3 to the third is equal to 27. 3 to the fourth is equal to 81. 3 to the fifth is going to be equal to-- this times 3 is 243. So it's going to be 243-- it's going to be some hairy math-- 243 over 5 minus-- well, 3 to the fourth is 81. Minus 81. Minus, let's see, we're going to have 3 to the third times-- let's see, so it's going to be minus 5 over 3 times 3 to the third power. So times 27. Well, 27 divided by 3 is just 9. 9 times 5 is 45. So let's just simplify that. So this is going to be equal to 45. Did I do that right? Yeah. It's essentially going to be like 3 squared times 5, because you're dividing by 3 here. So it's going to have 45. And then finally, 3 squared is 9. 9 times 12 is 108. So plus 108. These problems that involve hairy arithmetic are always the most stressful for me, but I'll try not to get too stressed. And then we're going to subtract out this whole thing evaluated at 0, but lucky for us, that's pretty simple. This evaluated is 0, 0, 0, 0, 0. So we're going to subtract out 0, which simplifies things a good bit. So now we just have to do some hairy fraction arithmetic. So let's do it. So what I'm going to do first is simplify all of this part. And then I'm going to worry about putting it over a denominator of 5. So let's see what we have. We have negative 81 minus 45. So these two right over here become negative 126. And then negative 126 plus 108. Well, that's just going to be the same thing as negative 26 plus 8, which is going to be negative 18. So this whole thing simplifies to negative 18. Did I do that right? So this is negative 126. And then negative 126-- yes. It will be negative 18. So now we just have to write negative 18 over 5. Negative 18 over 5 is the same thing as negative-- let's see, 5 times 10 is 50, plus 40. So that's going to be negative 90 over 5. So this whole thing has simplified to-- is equal to pi times 243 over 5 minus 90 over 5, which is equal to pi-- guys, I deserve a drum roll now. This was some hairy mathematics. So 243 minus 90 is going to be 153 over 5. Or we can write this as 153 pi over 5. And after all that math, you sometimes forget what we were doing in the first place. We were figuring out this right over here is the volume of this figure that had this little inside of it bored out.