5. GRAEME’S COMMENT OF WISDOM After all the Pi Day talk, we then drew our attention towards Graeme’s comment, which he had posted on Lawrence’s scribe post yesterday. Justus brought up the fact that its kind of like the triangle question we did two days ago. Of course, this one that Graeme has shown us is indeed a toughie. At the moment, I can’t quite wrap my head around it. I shall take another look over the weekend.

6. ROUND THE X-AXIS We then did somewhat of a review question, which asked to find the volume of the solid generated by f(x) and g(x) when revolved around the x-axis. All we had to do was follow the procedure of first taking one cross section and then taking a look at what we know about it.

7. ROUND THE X-AXIS From there, we then try to find the area of the washer by subtracting the small circle’s area from the big circle. We can then find the intersections of the two graphs so that we can find our interval to integrate between. Now integrate the area of the washer and you are done.

8. NOW THE Y-AXIS Now, we used the same functions, but instead we revolved around the y-axis. Like yesterday’s class, we can take apart the cylindrical shell so that it makes a rectangle. All we have to do from there is find the area of that rectangle, A = lwh. The height being the function, in this case, since we are looking for S, we’d just subtract the bottom function from the top one.

9. NOW THE Y-AXIS From there, we can now find our interval along which we will integrate by finding the intersections, as shown in Benchi’s little blurb at the corner there (He couldn’t talk so he had to explain the question by writing XD) After finding everything, just write your integral down and solve, which Mr. K didn’t care for us to do, just as long as we understood.

10. AROUND X = -3?! We then moved on from rotating around the y-axis to the more challenging x = -3. The first step would be to draw a diagram. That is key to helping you understand the question a bit more so that you can form your integral. As seen in the diagram, the radius would be the only problem in this question, since the cylindrical shell has gotten larger.

11. AROUND X = -3?! Of course, this would then mean that the hardest part of these questions would be figuring out the radius. The first thing we would do to find the radius in this problem is to find the part on the region S that would make the radius the smallest. This would be at x = 0.

12. AROUND X = -3?! Now if you take a look, you’d see that the radius is basically the distance from the rotation axis from the point x at which you are looking at. So from x = 0, the smallest radius you can get, you’d see that the radius is basically x + 3. From there, the procedure would be the same as before, just the fact that the radius is now x+3.

13. SAME SAME! Now, following the fact that “Math is the science of patterns”. We then revolved S around x = 6. Like I said in the previous slides, first find the point at which the radius will the smallest. Then take a look at the distance from the rotation axis. In this example, it would be 6-x.

14. FAVORITE NUMBER! Here, Mr. K asked us to pick our favorite number between 1 and 10. Although the majority of us had to tell him through sign language XD. Anyways, he was trying to introduce to us the average value theorem by first asking us how we would normally find averages. Which is to add up all the numbers and divide by the number of values we have.

15. AVERAGE VALUE FUN We then moved on and discussed how to find the average value of a function in the few minutes of class we had left. Wondering how this integral was derived? Well, the link shows exactly how with neat little animations. Go look at it, it will be the most worthwhile 2 minutes of your time 

16. HOMEWORK AND FAREWELLS Well, that’s all. I’m guessing that this slide is for homework since it was just left blank, along with Exercise 8.4 ODDS. The scribe for the weekend shall be Joseph. Do your homework okay or… Now I bid you all adieu. Till next time everybody :D