integration question :( (2 Viewers)

[lilkiwifruit]

I have my topic test soon and there are some integration questions where i keep getting the answer wrong, especially for ques that mentions "rotates around y axis" .

1.The region in the first quadrant bounded by the graphs of f(x)=1/8x^3 and g(x)= 2x is rotated about the y axis. Find the volume of the solid so formed.

Answer should be: 107.233 c. units

2. The curve x^2=16-4y^2 is rotated about the y axis. Find the volume of the generated solid.

Answer: 128pi/3 c. units

 

[lilkiwifruit]

Didn't you have it today? Anyway

nope

 

[pLuvia]

1.
y=1/8x³
y=2x
Limits y=0 and 8
int{0to8}pi[1/4y²-4y^2/3]dy
=pi[1/12y³-12/5y^5/3]{0to8}
=pi[0-(-512/15)]
=512/15pi u³

 

[pLuvia]

2.
Limits y=-2 and 2
2int{0to4}pi[16-4y²]dy
=2pi[16y-4/3y³]{0to2}
=2pi[0-(-64/3)]
=128/3pi u³

Edit: Answer only covered the top half of the ellipse

 

[pLuvia]

This is such a stupid ques but how did you get the limits 0 and 8 for the 1st ques?

Find the point of intersection in the first quadrant, and use the y coordinate of that intersection

 

[lilkiwifruit]

Find the point of intersection in the first quadrant, and use the y coordinate of that intersection

I get it now~! :rofl:

Are you sure that your answer for q 2 is right?? The answer in the textbook says 128pi/3 c. units.

 

[pLuvia]

Opps, ok editted the thing

 

[lilkiwifruit]

yep.. I understand now... Thank you~!

 

[lilkiwifruit]

I'm not trying to get you guys to do my hw for me but i really can't work this question out:

y= 2square root (1-x^2) and y=square root (1-x^2) intersect at (1,0) and (-1,0), find the volume enclosed in this region and rotated around 1) x axis and 2) the y axis.

Is this meant to be a semi-circle or something?

 

[hyparzero]

I'm not trying to get you guys to do my hw for me but i really can't work this question out:

y= 2square root (1-x^2) and y=square root (1-x^2) intersect at (1,0) and (-1,0), find the volume enclosed in this region and rotated around x axis and the y axis.

Is this meant to be a semi-circle or something?

its an ellipse with a semi-sphere of radius 1 cutout.

 

[Riviet]

I'm not trying to get you guys to do my hw for me but i really can't work this question out:

y= 2square root (1-x^2) and y=square root (1-x^2) intersect at (1,0) and (-1,0), find the volume enclosed in this region and rotated around 1) x axis and 2) the y axis.

Is this meant to be a semi-circle or something?

y=2sqrt(1-x²) is an ellipse, this turns out to be above the unit circle. So when finding the volume, subtract the volume formed by rotating circle from the volume formed by rotating the ellipse.

Let y₁=2sqrt(1-x²) and y₂=sqrt(1-x²)

V=∫pi[y₁² - y₂²] dx

Now substitute in your y's and integrate the rest. Hope that helps.

 

[Mc_Meaney]

Hijacking post - sorry.

Help, need to check my working and answers

1) Find d/dx of 2sinx+√x

d/dx= 2cosx+ 1/√x

2) ∫(cosx+5x²) dx

= sinx + 5x³/3

3) 0∫¹(1+e^-x) dx

= [x-e^-x]0¹
=(1-e^-1)-(e⁰
= |-0.368|
= 0.368units²

THanks in advance

 

[pLuvia]

1.
Answer should be
2cosx+1/2(x^-1/2)

2. Correct when
sinx+5x³/3+C

3.
Answer should be
2-e^-1

 

[Mc_Meaney]

thanks p. ill never get maths.

Edit : ok scrap that, how did u get the "2" in answer 3?

 

[SoulSearcher]

e⁰ = 1
(1-e^-1) - (0-e⁰) = 2 - e^-1

 

[Mc_Meaney]

thanks mate

 

[Mc_Meaney]

FInd the acceleration of P when t=1. THe equation is v= 1/t+1

How do i do thiss (not up to this topic fully)

 

[SoulSearcher]

I'm going to assume that v = 1/t + 1 is the velocity of the particle P
Acceleration is the derivative of velocity, so dv/dt = -1/t²
When t = 1, acceleration is -1/1² = -1/1 = -1 m/s²

 

[Mc_Meaney]

For displacement, would the eventual answer be 1/2?

 

[SoulSearcher]

I haven't got enough information to figure out the displacement of the particle P, a bit more information is needed to figure it out.