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# Washer method rotating around horizontal line (not x-axis), part 2

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.C (LO)
,
CHA‑5.C.4 (EK)
Doing some hairy algebra and arithmetic to evaluate the definite integral from the last video. Created by Sal Khan.

## Want to join the conversation?

• I might sound silly but why cant we just sink the figure into water and measure how much water fell of?
• That is not always practical. And, it is not, in real world practice, completely accurate. If you are making a part for a high performance engine, you may have tolerances of a few microns -- much too small to try to measure displaced water. Also, if the object in question is porous, it will absorb some of the water, thus invalidating the method.
• How would you find the area of an equation/equations that was rotated around a line that wasn't horizontal or vertical?
• Could you do U-substitution or something instead of multiplying the polynomial?
• He could use the identity a^2-b^2=(a+b)(a-b)
• If you were to revolve the functions about y=-4, would you find the same volume?
• 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)].
• When you raise x to the fourth at a little before and the result is 81, wouldn't that negative sign become positive? ie Yielding positive 81 in lieu of negative 81??
• 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.
• 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
and
(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
Meanwhile:
∫ (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.
• 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!
• 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).