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Calculating integral disc around vertical line

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.C (LO)
,
CHA‑5.C.2 (EK)
Calculate the integral from the last video. Created by Sal Khan.

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  • blobby green style avatar for user Henry Spivey
    Why is it that 5y is not integrated to 5y^2/2?
    (0 votes)
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  • leaf green style avatar for user Samuel Silva
    Does this shell has or not an empity core with radius 2. ?
    (sorry for the stupid question by the way,But this doubt is killing me.)
    (6 votes)
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    • piceratops ultimate style avatar for user Barrett Southworth
      This is the disc method of integration. The disc method can find the integral of a solid of revolution around an axis. It's finding the volume by pi*r^2*w, w = thickness of disc. When the volume is formed by revolving the equation about the y axis, the r is defined as a function of y. r = R(y) As integration is putting together ever smaller parts of it, pi*r^2*delta w as delta w approaches 0. So V = pi * integral of [R(y)]^2 dy, from a to b
      (2 votes)
  • blobby green style avatar for user Tahbee Hassan
    I think this is a washer method problem. in my understanding, we do washer's problem when a hole is created when the function is rotated to create inner and outer radius. By looking at this problem i thought there was an outer radius and inner radius. when you rotate the function around the y axis, the outer radius would be 2+x and the inner radius would be 2 because the distance of the inner radius is 2. so then the volume would be pi(outer radius)^2dy-pi(inner radius)^2dy? Is it wrong? Thanks
    (3 votes)
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  • aqualine sapling style avatar for user vidabehar
    where did the limits from -1 to 3 come from?
    (2 votes)
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  • piceratops ultimate style avatar for user TheAstronaut
    Is there a generalized equation for this somewhere?
    (2 votes)
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  • blobby green style avatar for user LynxDLL
    How do I interpret when the question says "enclosed by the x axis"? For example if a region is enclosed by the x axis and the curve y = (x^2) - 2x. What is the volume of the solid rotated around the x axis?

    What is the actual definition of enclosing? Because on this graph does it necessary mean the area under the x axis enclosed by the curve or the area above the x axis enclosed by the curve and the y axis? I'm just looking for a rigorous definition of enclosed by
    (2 votes)
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    • mr pink red style avatar for user andrewp18
      Your first interpretation is correct. Note that nowhere in the problem statement it was mentioned that the 𝑦-axis is a boundary. There is only one region that is bounded only by the 𝑥-axis and the curve. In general, only consider the region(s) bounded by the provided curves and/or axes.
      (1 vote)
  • leaf blue style avatar for user Jennifer
    Why is it that when you evaluate a definite integral with just u-substitution you have to re-calculate the values used for the bounds of integration, but in this case you can use u-substitution and keep your values the same?
    (1 vote)
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  • aqualine ultimate style avatar for user NathanielNeubert
    What if I wanted to calculate the volume of this solid using an x integral and using the shell form? What would my integral be and why?
    (1 vote)
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  • leafers ultimate style avatar for user leonferns96
    if we substituted back sqrt(y+1)=x and then evaluated our definite integral wouldn't it be much easier ??
    (1 vote)
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  • hopper cool style avatar for user Quantum Coding
    Can you rotate around a non vertical, non horizontal line? (like: y = x - 3) if so, could someone explain how?
    (1 vote)
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Video transcript

In the last video, we had set up the definite integral to evaluate the volume of this upside-down gumdrop truffle-looking thing. So now in this video, we can actually evaluate the definite integral. So what we need to do is really just expand out this expression, this square root of y plus 1 plus 2. So let's do that. So this is going to be equal to pi times the definite integral from y is equal to negative 1 to y is equal to 3. If you expand this out, you get square root of y plus 1 squared, which is just going to be y plus 1. And then you're going to have 2 times the product of both of these terms. 2 times 2 times square root of y plus 1 is going to be plus 4 times the square root of y plus 1. And then you have 2 squared, so plus 4. So you have this whole thing times dy. We can simplify a little bit. You have a 1 plus a 4. We can add the 1 to the 4 and get a 5. And now we're ready to take the antiderivative. So this is going to be equal to pi times-- let's take the antiderivative of all of this business-- pi times-- and I'll color code it. The antiderivative of y is just y squared over 2. The antiderivative of 4 times the square root of y plus 1-- you just really have to think of it as 4 times y plus 1 to the 1/2 power. We could use u substitution explicitly, but you probably are pretty practiced in this and can do this in your head. You have y plus 1 raised to the 1/2 power. Derivative of y plus 1 is just 1, which is essentially out here. So if you did u substitution, you would say u is equal to y plus 1. But this antiderivative is going to be equal to-- well, if you increment this exponent, you get 3/2 multiplied by the reciprocal 2/3. 2/3 times 4 is 8/3. So it's plus 8/3 times y plus 1 to the 3/2. And you can verify. If you take the derivative here you will get this expression right over here. 3/2 times 8/3 is 4. Decremented, you have y plus 1 to the 1/2 power. And then finally, you have-- let's see. What color have I not used yet? Finally, you have this 5. The antiderivative of 5 is just 5y. And we are going to evaluate it at 3 and at negative 1, y equals 3 and y equals negative 1. So this is going to be equal to pi. So let's evaluate all this business at 3. So 3 squared over 2 is 9/2. 3 plus 1 is 4 to the 3/2. Well, that's-- so let's see. If square root of 4 is 2 to the third power is 8, 8 times 8/3 is 64/3, so plus 64/3. You have 5 times 3. Well, that's going to be 15-- plus 15. And from that, we're going to subtract all this business evaluated at negative 1. So you have negative 1 squared over 2. Well, that's just 1/2. Negative 1 plus 1 is 0 to the 3/2 power. That's going to be 0 times 8/3. This is all going to be 0, so we don't have to even write it. And then finally, you have negative 1 times 5. Well, that's just going to be negative 5. And we are in the home stretch. We really just have to do a little bit of arithmetic, add some hairy fractions right over here. So let's do it. So this whole thing is going to simplify to pi times-- and it looks like-- let's see. Our least common multiple of all of these denominators is going to be 6. So let's put everything over a denominator of 6. So 9/2 is the same thing as 27/6. 64/3 is the same thing as 128/6. 15 is the same thing as 90/6. 1/2 is the same thing as 3/6. So you would distribute the negative sign. So this is negative 3/6. And negative times negative is positive. 5 is the same thing as 30/6, so plus 30/6. And so this is going to give us-- our denominator is going to be over 6. We're going to multiply something times pi. We have this pi over here. And then we just have to figure out what the numerator is. So let's see if I can do this in my head. So 27 plus 128 is going to be-- let me see. That's going to be 140, 155? Is that right? 155? Let's see, if we get to 48 plus another 7-- yeah, 155. Plus 90 gets us to 245. Is that right? Yeah. Plus 90 gets us to 245. Minus 3, you subtract 3 from that, you get to 242. And then you add 30 to that, you get to 272. So we're left with 272 pi/6. But then we can-- let's see. 272 and 6 are both divisible by 2. So this is equal to-- let's see. 272 divided by 2 is going to be 136 pi over-- and if you divide this denominator right over here by 2-- over 3. Is that right? Yeah. 136 pi/3. And 136 is not divisible by 3. So we have it as simplified as we can. This right over here is the volume of our little upside-down gumdrop-looking thing.