## anonymous one year ago rotation of a bounded area question i can't solve Write the integral in one variable to find the volume of the solid obtained by rotating the first-quadrant region bounded by y = 0.5x^2 and y = x about the line x = 5.

1. anonymous

my trouble is mainly with the fact that the rotation is done around x=5 and not the x axis

2. anonymous

im sure something changes but not sure what

3. anonymous

Do you prefer the washer or shell method?

4. anonymous

well i don't know the shell method so washer i suppose

5. anonymous

Observe that the dimensions of the washer that you'll be interested in (i.e. radius) will have to be written in terms of $$y$$. Let's draw the region first: |dw:1439519332453:dw|

6. anonymous

so does that mean we will have to integrate with sespect to y?

7. anonymous

Yes! Since $$y=\dfrac{1}{2}x^2$$, you have $$x=\sqrt{2y}$$ (taking the positive root since we're in the first quadrant). Here's a washer associated with an matching of points $$(x,y)=(\sqrt{2y},y)$$ and $$(x,y)=(y,y)$$ on the boundary of the region: |dw:1439519657597:dw|

8. anonymous

(As you'll no doubt learn soon enough, the shell method allows you to integrate with respect to $$x$$ instead.)

9. anonymous

Correction: the inner radius should be $$5-\color{red}{\sqrt{2y}}$$

10. anonymous

The volume of the solid is the infinite sum of the areas of these infinitely thin washers. The area of a washer is essentially the difference of the areas of circles. Take some circle with radius $$OR$$ and put some smaller circle with radius $$IR$$ at the larger one's center. The area of the large circle is $$\pi(OR)^2$$, and that of the smaller circle is $$\pi(IR)^2$$. Remove the smaller circle from the larger, and you're left with the washer shape, which has area $$\pi((OR)^2-(IR)^2)$$. So, to add up all these areas, you take the integral. It'll look something like this (I'll leave the details to you): $V=\int_a^b \pi((OR)^2-(IR)^2)\,dy$

11. anonymous

okay sorry lost internet connection , back now give me a second to read all that you posted

12. anonymous

okay so tell me if this is way off $V = \pi \int\limits_{0}^{2}(y^2)-({2y}) dy$

13. anonymous

It's off by little bit. Your setup would be right if you were revolving about the $$y$$-axis (the line $$x=0$$).

14. anonymous

okay so do i just add 5 to each expression or something?

15. anonymous

Sort of... if you take another look at that sketch, you'll notice that I lay out the lengths of a few distances. Let me remove some of that clutter: |dw:1439520860137:dw| Notice that the two smaller distances must add up to $$5$$. That is, $y+OR=5$ This means $$OR=5-y$$, not just $$y$$.

16. anonymous

Similarly, you can figure out what the inner radius would be by breaking down these distances: |dw:1439521024630:dw| You have $$\sqrt{2y}+IR=5$$, so $$IR=5-\sqrt{2y}$$.

17. anonymous

okay thats interesting let me give it another shot $V = \pi \int\limits_{0}^{2} (-x-5)^2 - (5-\sqrt{2y}) dy$

18. anonymous

woops forgot to square the last part

19. anonymous

I think you meant $$5-x$$, or $$-x+5$$ for the first term? Squared, of course.

20. anonymous

$$y$$, not $$x$$**

21. anonymous

22. anonymous

im getting 5.333 , does that look right to you?

23. anonymous

Yup

24. anonymous

times $$\pi$$, right?

25. anonymous

oh yea forgot :)

26. anonymous

It doesn't look like you actually had to do the computation, but maybe those instructions were just left out? Anyway, setting up these types of integrals is the hardest part.

27. anonymous

(usually)

28. anonymous

thanks for the help ! :D

29. anonymous

You're welcome!