Thread: In need of verification and help, integration of volumes

Hi. I am not sure if these questions have been posted before. Reworked them today, the answers are not in the book:

Find volume generated by revolving the regions bounded by given curves about y axis
Question 26-3-20
y^2 = x, y = 4, x = 0, Disks

Attempt:
dV = pix^2
x^2=y^4
Integration: pi(1/5)(4^5) = 634.4 ?

Question 26-3-22
y = 3x^2 - x^3, y = 0, Disks

Attempt:
dV = pix^2
(3x^2 - x^3)(3x^2 - x^3) =
6x^4 - 6x^5 + x^6
Intercepts: x = 3 when y = 0
Integration:
pi 6(3^4) - 6(3^5) + 3^6 =
pi (486-1458+729) = -763.4 ?
negative answer is not good

Question 26-3-24
x^2 + 4y^2 = 4, Disks

Attempt:
Isolating x: x^2 = 4 - 4y^2
x^2 = 4y - 1/3(4(y^3)) ?
Please let me know if this is correct so far.

Thanks for your time.

Hey togo.

Q1 looks right.

For Q2, what are the limits? (Are you bounding the area by some other function?)

For Q3, try setting up your limits (between -1 and +1) and see you how go.

All the info I have has been posted, for Q2 I don't understand why no limit was given either.

For Q2 you didn't calculate the anti-derivative properly: it should be:

F(x) = 6x^5/5 - x^6 + x^7/7

This gives an integral value of 6/5*3^6 - 3^6 + 3^7/7 =

> 6/5*3^6 - 3^6 + 3^7/7
[1] 458.2286

Can you try Q3?

Tried Q3, the question also says Quadrant 1.
I was looking for x intercepts in order to find y boundary values, but end up dividing by zero.

y = 1 - (x^2)/4

Show us all of your working.

Sorry the original equation I gave (y = 1 - (x^2)/4) was wrong (needs a square root)

4y^2 = 4 - x^2
y^2 = 1 - (x^2)/4
y = (1 - x^2/4)^(1/2)

is just a reworking of the problem equation algebraically in order to find the x intercepts and therefore find the y boundary to plug them into the integral.

the integral which I cannot even try using yet is x^2 = 4y - 1/3(4y^3) although not proven correct.

no boundaries were given in the question except "quadrant 1"

If its quadrant 1, then your integral will be bound to x = -2 and x = 0.

Using this hint, can you evaluate the integral?

Did you graph the function in problem 2? If you do you will see what region it is that is rotated around the y-axis. You will need to do two "disk method" integrations.

damn. (Q3, working on Q2 soon)

This is what I have:
x^2 = 4y - 1/3(4y^3)

principle formula dV = pix^2

Tempted to plug -2 into y in the above integral
but we plug in x as y?
4(-2) - 1/3(4(-2^3))?

I still get a weird answer for integral on Q2

6x^4 - 6x^5 + x^6
1/5(6x^5) - 1/6(6x^6) + 1/7(x^7)
291.6 - 729 + 312.43 = weird negative answer!

If you were to graph , you would see that there is a loop from (0, 0) up to (2, 4) and down to (3, 0). Rotating that around the y-axis gives a "doughnut" shaped figure- there is a hole in the middle. If you are using the "shell method" (though you said in your first post "Disks"), the radius of a disk would be x and the height y so you would have .

If you really want to use the "disk" method the radius should be x in terms of y and I don't see any good method of solving for x.

Yes shell method, not sure how disks got in there. I plotted it out and can see it now. Although I don't see any hole in the middle