2 Questions (1 Viewer)

[Dreamerish*~]

1. Find the coordinates of the point P which divides the interval AB internally in the ratio 2:3 where A and B have coordinates (1, -3) and (6, 7) respectively.

i keep getting P(4, 3) but the answer should be P(3, 1)

2. Find the volume of the solid obtained when the region between the curves y = x³ and y = x², from x = 0 to x = 1, is rotated about the x axis.

i wasn't 100% about how to do this one. i did the "roof - floor" thing and squared it, and i got π/105. the answer is 2π/35

 

[KFunk]

1. Find the coordinates of the point P which divides the interval AB internally in the ratio 2:3 where A and B have coordinates (1, -3) and (6, 7) respectively.

i keep getting P(4, 3) but the answer should be P(3, 1)

2. Find the volume of the solid obtained when the region between the curves y = x³ and y = x², from x = 0 to x = 1, is rotated about the x axis.

i wasn't 100% about how to do this one. i did the "roof - floor" thing and squared it, and i got π/105. the answer is 2π/35

For the first one, remember the general formula for internal division of an interval:
When dividing the points (x₁,y₁) and (x₂,y₂) into the ration m:n the coordinate is:

(mx₂ +nx₁)/(m+n) , (my₂ +ny₁)/(m+n)

using your points and ratio:

[2(6) +3(1)]/5 , [2(7) - 3(3)]/5 = (3,1)

 

[KFunk]

For the second question remember that if 0 < x < 1 then x³ < x² hence you need to subtract the volume of the rotation of y=x³ from y=x²

Assume that all the following integrals are between 0 and 1:

V = &pi;&int; x⁴ dx - &pi;&int; x⁶ dx
= &pi;&int; x⁴ - x⁶ dx
= &pi;[ 1/5.x⁵ - 1/7.x⁷]
= &pi;[1/5 - 1/7]
= &pi;[ 2/35]
= 2&pi;/35

 

[tempco]

btw, how do you do that cool integration thingy?

 

[KFunk]

If you click 'quote' on my post you should be able to see how. Otherwise type:

& ----then---> int ---then---> ; without any spaces.

Check out this post by Xayma:

http://www.boredofstudies.org/community/showthread.php?t=7127&page=3

 

[KFunk]

i wasn't 100% about how to do this one. i did the "roof - floor" thing and squared it, and i got π/105. the answer is 2π/35

For the method you used what you've done is take the distance between y=x² and y=x³ and rotated that around the axis. What that doesn't take into account is the distance the strips are from the axis of rotation:

Imagine a strip one unit high from (1,0) to (1,1).
When revolved around the x-axis you get a circle with the area &pi;r² = &pi; units²

If you raise this strip so that it sits between (1,1) and (1,2) and then revolve it around the x-axis then things change. The outer point of the strip is tracing out a circle of radius (r) = 2.
Hence the area of the ring that the strip traces = &pi;r₁² - &pi;r₂² = 4&pi; - &pi; = 3&pi; units²

I hope that makes sense of why your original method didn't give you the right answer.

 

[Dreamerish*~]

For the method you used what you've done is take the distance between y=x² and y=x³ and rotated that around the axis. What that doesn't take into account is the distance the strips are from the axis of rotation:

Imagine a strip one unit high from (1,0) to (1,1).
When revolved around the x-axis you get a circle with the area &pi;r² = &pi; units²

If you raise this strip so that it sits between (1,1) and (1,2) and then revolve it around the x-axis then things change. The outer point of the strip is tracing out a circle of radius (r) = 2.
Hence the area of the ring that the strip traces = &pi;r₁² - &pi;r₂² = 4&pi; - &pi; = 3&pi; units²

I hope that makes sense of why your original method didn't give you the right answer.

yeah i was kind of half-guessing that one.
lol the first one was pretty stupid. '-_- somehow i didn't see that it's mx₂ and i thought it was mx₁

thanks for your help xD