integration help (1 Viewer)

[Halo189]

1) find the exact area bounded by the parabola y=x^2 and the line y=4-x
2) find the volume of the solid formed when the curve y=(x+5)^2 is rotated about the y-axis from y=1 and y=4

 

[snowflakeptical]

1) find the exact area bounded by the parabola y=x^2 and the line y=4-x
2) find the volume of the solid formed when the curve y=(x+5)^2 is rotated about the y-axis from y=1 and y=4

1)

 

[Carrotsticks]

1)

these are composite areas, so split it up and find pts if intersection
find area:

integrated you get
[4x - x^2/2 - x^3/3]
sub in values
[4(2)-(2)^2/2 -(2)^3/3]-0 + [0- (4(-2)-(-2)^2/2-(-2)^3/3]
=10/3 + 26/3
=36/3 =12

getting to 2)

You can find the answer to the 1st question by integrating from -2 to 4 directly in one integral, as the top and bottom curves are the same over that domain =)

You'd only need to split it into two integrals if the curves change.

 

[snowflakeptical]

You can find the answer to the 1st question by integrating from -2 to 4 directly in one integral, as the top and bottom curves are the same over that domain =)

You'd only need to split it into two integrals if the curves change.

oh yeah true
Does the answer still work?

Hang on, I got this


wait, why did carrot leave

 

[Halo189]

The answer is apparently (17 sqrt17)/6 units^2

 

[InteGrand]

The limits of the integral shouldn't be -2 and 2.

The limits should be where 4 – x = x², which is when .

 

[snowflakeptical]

The limits of the integral shouldn't be -2 and 2.

The limits should be where 4 – x = x², which is when .

integrated is [4x-x^2/2 -x^3/3] then sub in values?

but yeah I lost the plot

 

[InteGrand]

integrated is [4x-x^2/2 -x^3/3] then sub in values?

but yeah I lost the plot

Yeah it's just a bit tedious.

 

[snowflakeptical]

then evaluate

I keep getting it wrong though

 

[Silly Sausage]

The integral should be:

 

[snowflakeptical]

Did you expand the squares and cubes correctly? (The formula to expand cubes is .)

there is no easier way is there?


------------------------------------------------
Q2.
y=(x-5)^2
rearrange to make x the subject
√y = x+5
x=√y -5
vol= pi ∫( √y - 5)^2 dy
integrated:
pi[y^2/2 - 5y^3/2 + 25y]
then sub in values

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scroll down to the answer

 

[Halo189]

Dont worry everyone, I understood it, but the working out to long

 

[snowflakeptical]

Dont worry everyone, I understood it, but the working out to long

yeah very :/

 

[Carrotsticks]

Those limits are yuck.

 

[Halo189]

Thanks for your help

 

[snowflakeptical]

Those limits are yuck.

would they make us do a q like this in an exam. If so, how many marks worth?

Thanks for your help

=)
btw, is the answer to q2 correct?
85pi units^3

 

[Halo189]

would they make us do a q like this in an exam. If so, how many marks worth?

=)
btw, is the answer to q2 correct?
85pi units^3

I dont think so because its way too long
And that question is also confusing... answer is apparently 215pi/6
Its fine my friends and I are working on this exercise tomorrow

 

[Carrotsticks]

would they make us do a q like this in an exam. If so, how many marks worth?

=)
btw, is the answer to q2 correct?
85pi units^3

The type of Q (finding the area of the region bounded by a curve and a line) has been done MANY times in examinations!

But the ugly limits... rarely the case. You will usually have nice limits =)

If the question was just "Find the area of ....", then it would have to be worth 3 marks, possibly 4 if they are generous. One for finding the intersections, one for obtaining the correct integral and one for obtaining the correct answer.

 

[Silly Sausage]