finding area enclosed by 2 curves

[Ineptskills]

Hi,

Please see my work so far

the curves are y=x3 and y=4x-x^2

I'm not sure if I have gone wrong somewhere? maybe from the beginning

Thanks

 

[Deleted member 4993]

Hi,

Please see my work so far

the curves are y=x3 and y=4x-x^2

I'm not sure if I have gone wrong somewhere? maybe from the beginning

Thanks

Did you "approximately" sketch the curves?

That is the first thing to do in these problems.

 

[Ineptskills]

Did you "approximately" sketch the curves?

That is the first thing to do in these problems.

I tried but the graph of y=x^3 is almost a horizontal line when compared to the graph of y=4x-x^ is that ok or is there a way around it?

 

[Amoren]

The math course I'm going through recently taught how to do problems like this with the use of double integrals, is this something along the lines of how they're expecting you to solve the problem as well? If so, I could detail one of the examples for a similar problem (finding the enclosed area between the curves x^2 and 8 - x^2 ) to help get you going on the right track if need be.

 

[Ineptskills]

The math course I'm going through recently taught how to do problems like this with the use of double integrals, is this something along the lines of how they're expecting you to solve the problem as well? If so, I could detail one of the examples for a similar problem (finding the enclosed area between the curves x^2 and 8 - x^2 ) to help get you going on the right track if need be.

That would be extremely helpful, I'm struggling to get calculus

All the question states is [find the area enclosed by the curves: y=x^3 and y=4x-x^2]

My graph is kind of crude, not sure if it's correct

 

[Amoren]

Alright, I don't have a camera or a phone, so I can't take pictures of neat written out formulas, forgive what I can get across with typing.

I'll be going through an example of the area enclosed by the curves 8 - x^2 and x^2.

First, you need to find the domain and range for the enclosed area you're trying to find the area for. Essentially, what values of x and y it can have and still be considered in the area. Its easiest to write these as an inequality in relation to x or y.

By setting 8 - x^2 and x^2 equal to each other, we find that the points where our two curves meet are -2 and 2, which form the farthest points of x of our enclosed area, or -2 <= x <= 2. X can be any point between -2 and 2 and still have a point in the area. Y is trickier, as the values for it are determined by the actual curves. x^2 <= y <= 8 - x^2, Y can be any value within or on these two curves for any point of x.

We then take a double integral, bounded by the domain and ranges of our x and y value, and apply it to the integrand 1 (if we were trying to find a volume, or other such three-dimensional equation, we'd have an integrand that wasn't one). Ergo, integrand(-2 to 2) dx * integrand(x^2 to 8 - x^2)dy, more appropriately written as integrate(-2 to 2) * integrate(x^2 to 8 - x^2)dA (where dA = dx*dy).

We start with our y term, since it is closest to our integrand, 1. The integral of 1 with respect to y is y. We then solve for the bounds, getting [8 - x^2] - [x^2] (substituting 8 - x^2 and x^2 as if they were normal values for y like any typical bounded integral), leaving us with simplified) integrate(-2 to 2) 8 - 2x^2 dx. The integral of 8 - 2x^2 with respect to x is 8x - 2(x^3)/3, then solving for the bounds and substituting in -2 and 2, we get; [8*2 - 2(2^3)/3] - [8*-2 - 2(-2)^3/3]. Simplified as [16 - 16/3] - [-16 + 16/3], or 32 - 32/3, or 64/3 as the area of our curve.

This is assuming I haven't screwed up any of the calculations, I'm busy at work on my own assignment due in four hours.

 

[Steven G]

Hi,

Please see my work so far

the curves are y=x3 and y=4x-x^2

I'm not sure if I have gone wrong somewhere? maybe from the beginning

Thanks

You have to know algebra and arithmetic before you can do calculus. The calculus part of calculus is actually quite easy. After all, most of the times you are taking the derivative or computing the integral of polynomials. But it is the algebra that beats most students. You are having trouble dealing with a subtraction in your problem. That "-x" should be +x. You do not know how to find where two functions intersect. You do not know how to sketch a function. But notice that you were able to compute the integral correctly, which is calculus!
I am not trying to give you a hard but before you attempt to study calculus you just have to know algebra and arithmetic. Sit in on an algebra class, study from a book, watch videos, but learn algebra.

 

[Ineptskills]

Alright, I don't have a camera or a phone, so I can't take pictures of neat written out formulas, forgive what I can get across with typing.

I'll be going through an example of the area enclosed by the curves 8 - x^2 and x^2.

First, you need to find the domain and range for the enclosed area you're trying to find the area for. Essentially, what values of x and y it can have and still be considered in the area. Its easiest to write these as an inequality in relation to x or y.

By setting 8 - x^2 and x^2 equal to each other, we find that the points where our two curves meet are -2 and 2, which form the farthest points of x of our enclosed area, or -2 <= x <= 2. X can be any point between -2 and 2 and still have a point in the area. Y is trickier, as the values for it are determined by the actual curves. x^2 <= y <= 8 - x^2, Y can be any value within or on these two curves for any point of x.

We then take a double integral, bounded by the domain and ranges of our x and y value, and apply it to the integrand 1 (if we were trying to find a volume, or other such three-dimensional equation, we'd have an integrand that wasn't one). Ergo, integrand(-2 to 2) dx * integrand(x^2 to 8 - x^2)dy, more appropriately written as integrate(-2 to 2) * integrate(x^2 to 8 - x^2)dA (where dA = dx*dy).

We start with our y term, since it is closest to our integrand, 1. The integral of 1 with respect to y is y. We then solve for the bounds, getting [8 - x^2] - [x^2] (substituting 8 - x^2 and x^2 as if they were normal values for y like any typical bounded integral), leaving us with simplified) integrate(-2 to 2) 8 - 2x^2 dx. The integral of 8 - 2x^2 with respect to x is 8x - 2(x^3)/3, then solving for the bounds and substituting in -2 and 2, we get; [8*2 - 2(2^3)/3] - [8*-2 - 2(-2)^3/3]. Simplified as [16 - 16/3] - [-16 + 16/3], or 32 - 32/3, or 64/3 as the area of our curve.

This is assuming I haven't screwed up any of the calculations, I'm busy at work on my own assignment due in four hours.

Thanks so much, you really shouldn't have. I will refer to this extensively as I can't seem to follow the vids on youtube as they don't explain as well as this as you have described your thinking as well as what would be written.

Really, thanks very much Amoren :cool:

 

[Steven G]

Alright, I don't have a camera or a phone, so I can't take pictures of neat written out formulas, forgive what I can get across with typing.

I'll be going through an example of the area enclosed by the curves 8 - x^2 and x^2.

First, you need to find the domain and range for the enclosed area you're trying to find the area for. Essentially, what values of x and y it can have and still be considered in the area. Its easiest to write these as an inequality in relation to x or y.

By setting 8 - x^2 and x^2 equal to each other, we find that the points where our two curves meet are -2 and 2, which form the farthest points of x of our enclosed area, or -2 <= x <= 2. X can be any point between -2 and 2 and still have a point in the area. Y is trickier, as the values for it are determined by the actual curves. x^2 <= y <= 8 - x^2, Y can be any value within or on these two curves for any point of x.

We then take a double integral, bounded by the domain and ranges of our x and y value, and apply it to the integrand 1 (if we were trying to find a volume, or other such three-dimensional equation, we'd have an integrand that wasn't one). Ergo, integrand(-2 to 2) dx * integrand(x^2 to 8 - x^2)dy, more appropriately written as integrate(-2 to 2) * integrate(x^2 to 8 - x^2)dA (where dA = dx*dy).

We start with our y term, since it is closest to our integrand, 1. The integral of 1 with respect to y is y. We then solve for the bounds, getting [8 - x^2] - [x^2] (substituting 8 - x^2 and x^2 as if they were normal values for y like any typical bounded integral), leaving us with simplified) integrate(-2 to 2) 8 - 2x^2 dx. The integral of 8 - 2x^2 with respect to x is 8x - 2(x^3)/3, then solving for the bounds and substituting in -2 and 2, we get; [8*2 - 2(2^3)/3] - [8*-2 - 2(-2)^3/3]. Simplified as [16 - 16/3] - [-16 + 16/3], or 32 - 32/3, or 64/3 as the area of our curve.

This is assuming I haven't screwed up any of the calculations, I'm busy at work on my own assignment due in four hours.

I am sorry but this us not correct. Clearly x=0 is a point of intersection and x=2 and x=-2 are not points of intersection. Also, I suspect that this student is in calculus 1 and does not know about double integrals.

 

[Steven G]

Hi,

Please see my work so far

the curves are y=x3 and y=4x-x^2

I'm not sure if I have gone wrong somewhere? maybe from the beginning

Thanks

Step 1: You need to find the x-values of the points of intersection.
X^3 = 4x-x^2, x^3+x^2-4x=0, x(x^2 + x - 4)=0, so x=0, x=(-1+sqrt17)/2 and x= (-1-sqrt17)/2.
Step 2: order the three x-values: (-1-sqrt17)/2, 0, (-1+sqrt17)/2.
Step 3: Between (-1-sqrt17)/2 and 0 we need to know which graph is on top.
Between 0 and (-1+sqrt17)/2 we need to know which graph is on top.
NOTE: For each of these two intervals only one graph will be on top over the whole interval. If this was not true, that is for part of the interval one graph was on top then for the remainder of the interval the other graph was on top then the graphs would intersect!
(-1-sqrt17)/2 and 0: sqrt17 is a little more than sqrt16 which is 4. So (-1-sqrt17)/2 is a little below -5/2. Try x=-1: for y1=x^3 we get y =(-1)^3=-1. For y2=4x-x^2, we get y2= 4(-1)-(-1)^2 = -4-1=-5. So y1>y2 on this interval.
(-1+sqrt17)/2 and 0: (-1+sqrt17)/2>3/2. So try x=1. y1=1^3=1 and y2=4(1)-(1)^2=3 so y2>y1 on this interval.
Step 4: We integrate. But we always integrate the top function - bottom function. Since the top graph changes we will have multiple integrals.
Area= integral from (-1-sqrt17)/2 to 0 of [(x^3)-(4x-x^2)]dx + integral from 0 to (-1+sqrt17)/2 of [(4x-x^2)-(x^3)]dx. Continue from here.

You can skip step 2 if you like as long as you take the absolute value of each integral, that is you can have y1-y2 or y2-y1 for each integral

 

[Amoren]

I am sorry but this us not correct. Clearly x=0 is a point of intersection and x=2 and x=-2 are not points of intersection. Also, I suspect that this student is in calculus 1 and does not know about double integrals.

Note that I wasn't doing his problem, but going into a different problem with the techniques my course has taught me to solve it. X=2/-2 is the points of intersection for x^2 and 8-x^2, and its certainly not 0 (which would give y values of 0 and 8, respectively.)