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.
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.
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?
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"
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))?
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.