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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition)>Unit 11

Lesson 6: Volume: disc method

# Calculating integral disc around vertical line

Calculate the integral from the last video. Created by Sal Khan.

## Want to join the conversation?

• Why is it that 5y is not integrated to 5y^2/2?
• Ok, I'm a little (read: a lot!) confused by these previous answers, but you are right, the integral of 5y would be 5(y^2)/2 — but look again! We weren't taking the integral of 5y, we were taking the integral of 5!
• 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.)
• 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
(1 vote)
• 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
• That would be true, if he had cut the distance between the y axis and x=2 out of the function, but he chose not too and left that whole cylindrical shape inside. TLDR he didn't cut that part out of the shape.
• where did the limits from -1 to 3 come from?
• In the last video Sal defined the region that we care about (want to evaluate the volume of) to be from y = -1 up to y = 3.
• Is there a generalized equation for this somewhere?
• 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
• 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)
• 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)
• We plugged back in the y+1, that is why the bounds didn't change. If you would have kept the u, then you would have to recalculate the bounds reflecting that.
• 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)
• if we substituted back sqrt(y+1)=x and then evaluated our definite integral wouldn't it be much easier ??
(1 vote)
• Not sure I understand the question, but I tried using y = x^2 from 0 to 4 and got the same result with easier integration and evaluation as you suggest.
(1 vote)
• Can you rotate around a non vertical, non horizontal line? (like: y = x - 3) if so, could someone explain how?
(1 vote)
• One more method would be to rotate the coordinate axes so that the given line is now parallel to one of them and then rotating about these new axes.
(1 vote)