# Area between curves/solids of revolutions

1. Apr 29, 2007

### t_n_p

I'll just ask one question at a time as I don't want to spam up the forum!

1. The problem statement, all variables and given/known data
Find the area of the region enclosed by the curves y²=4+x and x+2y=4. 2.

3. The attempt at a solution
Well I know I have to transpose the two equations to make y the subject, but for the first equation I'm unsure what to do with the square root. Do I take it as a circle (i.e. ±root[4+x]) ?

2. Apr 30, 2007

### gunstar-red

First of all make y the subject so your two equations will be
$$x = y^2-4 \\ x= 4-2y$$
Next equate the two equations to find the points of intersection
$$\begin{eqnarray*} y^2-4 &=& 4-2y\\ y^2 +2y -8 &=& 0\\ (y-2)(y+4) &=& 0 \end{eqnarray*}$$

y = 2 or -4

These will be the upper and lower limits. We then integrate with respect to y not with respect to x so as to avoid the ±root[4+x].
$$Area = \int_{-4}^2 (4-2y)-(y^2-4) dy\\ =\int_{-4}^2 8-2y-y^2 dy\\ =[8y-y^2-\frac{y^3}{3}]_{-4}^{2}\\ =36$$

Last edited: Apr 30, 2007
3. Apr 30, 2007

### t_n_p

Ah thanks, so much easier!

4. Apr 30, 2007

### t_n_p

I've got another question: Find the volume of the solid of revolution generated by rotating about the y axis and the region enclosed by the y-axis and the curves y=2x²-1 and y=root(x).

I know they intersect at x=1 and y=root(x) forms the "top" part if your looking at it. But because we are rotating about the y-axis, does that now mean that y=2x²-1 forms the "top".

Just in case you don't know what I mean by "top", when you have area inclosed you do top-bottom.

5. Apr 30, 2007

### HallsofIvy

Staff Emeritus
No, the way you are rotating has nothing to do with what is the "top". One thing you need to clarify is whether you want to do this problem by "shells" or "washers".

6. May 1, 2007

### t_n_p

I'm familiar with the term shells, however I'm not sure what you mean by washers. It's just the standard volume of a revolution. I was taught to do it like this method:

7. May 2, 2007

### t_n_p

*bump*
anybody?

8. May 2, 2007

Last edited: May 2, 2007
9. May 2, 2007

no, rt(x) is the 'top'

draw a picture of rt(x) and 2x^2-1 you can see easily which one is the 'top'

10. May 3, 2007

### t_n_p

Isnt the rule V=pi*∫ x² dy?

how come in the example i posted there is 2*pi*x ?

11. May 3, 2007

As mentioned by HallsofIvy, you need to decide whether you would like to use the washer or shell method to solve the problem. For this question, both methods are fine.
If you use the washer method, you'll have V = pi * ∫ x² dy.
Otherwise, for the shell method, you'll get V = 2 * pi * ∫ xy dx.
Note that you'll need to modify these formulas if you are rotating about the x-axis.

For more details on the washer method, refer to http://en.wikipedia.org/wiki/Disk_integration
For more details on the shell method, refer to http://en.wikipedia.org/wiki/Shell_integration

Last edited: May 3, 2007
12. May 4, 2007

### t_n_p

Thanks, decided to use the shell method and got my final answer to be 4pi/5 cubic units. Would somebody be able to verify this for me?

13. May 4, 2007

The question: Find the volume of the solid of revolution generated by rotating about the y-axis the region enclosed by the x-axis and the curves $$y=2x^2-1$$ and $$y=\sqrt{x}$$.

No, I don't think I got the same answer as you.

If you used the shell method to solve this problem, you would need to take the difference of 2 definite integrals to obtain the final answer. Did you get something like this...

$$V = 2\pi \int x \sqrt x dx - 2 \pi \int x(2x^2-1)dx$$

Can you figure out what the limits of the two integrals should be? Note that they are not the same! (For this reason, the washer method may be slightly easier, because the limits of the two integrals will be the same. Try using the washer method and see!)

Last edited: May 4, 2007
14. May 4, 2007

### t_n_p

I can't find where ive gone wrong in the shell method...

If I use washer method, what will go in the place of "x²" ?

15. May 4, 2007

### gunstar-red

I got 4pi/5 cubic units, I'm pretty sure you're right.

16. May 5, 2007

### t_n_p

hooray! :rofl:

17. May 6, 2007

I'm not sure if I'm right, but this is what I got:-

$$V = 2\pi \int_{0}^{1} x \sqrt x dx - 2 \pi \int_{\frac{1}{\sqrt{2}}}^{1} x(2x^2-1)dx$$

This will eventually give me $$\frac{4\pi}{5}-\frac{\pi}{4}$$, which is the same as $$\frac{11\pi}{20}$$ cubic units.

Last edited: May 6, 2007
18. May 6, 2007

### t_n_p

Why is your lower limit for the second integral 1/root(2) ?