# Calculating the surface area of a solid of rotation

• Wi_N
In summary, according to Wolfram Alpha, the function does not have a nice form and no solution exists.
Wi_N
Homework Statement
Calculate surface area of the solid when a curve is rotated around x axis
Relevant Equations
x^(a/b) + y^(c/d) = 1
For some reason I have become very unsure but my gut feeling says i can calculate y=(1-x^(a/b))^(d/c)
I already know the formula for calculating the volume. but can transfer the whole thing as a function of y(x) and take the integral then as a single integral?

What is it you want to calculate ? Is there a complete problem statement ? Volume or area ?

Wi_N
BvU said:
What is it you want to calculate ? Is there a complete problem statement ? Volume or area ?

the area of x^(3/2) + y^(3/2) =1 rotating around x-axis.

can i rewrite the equation as a function y(x).

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By all means !
Did you have bounds for the integration in mind ?

BvU said:
By all means !
Did you have bounds for the integration in mind ?

obviously x is between 1 and 0 since those are the max and min values looking at the original equation.
my plan is to rewrite it as f(x) and then use the standard area integral formula to solve it. using boundries of 0 and 1. is this correct?

Yes

Wi_N
Wi_N said:
Homework Statement: Calculate volume of integral.
Homework Equations: x^(a/b) + y^(c/d) = 1

For some reason I have become very unsure but my gut feeling says i can calculate y=(1-x^(a/b))^(d/c)
I already know the formula for calculating the volume. but can transfer the whole thing as a function of y(x) and take the integral then as a single integral?
In the future, please take care to provide a complete problem statement. In this case, it would have been helpful to include the information that the region bounded by the graph of the equation and the x- and y-axes is revolved around the x-axis.

Wi_N
So, @Wi_N , did you come up with an integral ?
I'n not very good with those , so all I could think of is a numerical solution ...

BvU said:
So, @Wi_N , did you come up with an integral ?
I'n not very good with those , so all I could think of is a numerical solution ...

##S=2\pi\int_{0}^{1} (1-x^{\frac{3}{2}})^{\frac{2}{3}})\sqrt{1-(-\frac{\sqrt{x}}{\sqrt[3]{1-x^\frac{3}{2}}})^2}.##

haven't solved it yet though. but this is the easy part. what does your numerical solution look like?

edit: this integral seems very problematic. thinking its wrongly written perhaps...starting to think a parameter change is needed..

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$$S=2\pi\int_{0}^{1} (1-x^{\frac{3}{2}})^{\frac{2}{3}})\sqrt{1-(-\frac{\sqrt{x}}{\sqrt[3]{1-x^\frac{3}{2}}})^2}.$$

(you want to use double $signs to enclose displayed math) I don't recognize the formula for volume of solid of revolution , i.e. ##\pi \displaystyle \int \bigl ( y(x) \bigr )^2 \;dx\ ## with, in your case ## y=\Bigl (1-x^{3\over 2} \Bigr)^{2\over 3} ## . BvU said: $$S=2\pi\int_{0}^{1} (1-x^{\frac{3}{2}})^{\frac{2}{3}})\sqrt{1-(-\frac{\sqrt{x}}{\sqrt[3]{1-x^\frac{3}{2}}})^2}.$$ (you want to use double$ signs to enclose displayed math)

I don't recognize the formula for volume of solid of revolution , i.e. ##\pi \displaystyle \int \bigl ( y(x) \bigr )^2 \;dx\ ## with, in your case ## y=\Bigl (1-x^{3\over 2} \Bigr)^{2\over 3} ## .

thanks, it was for area, not volume. and it was not correct since its int fx * sqrt(1+(f'x)^2) still I am not getting anything i can solve..

$$S=2\pi\int_{0}^{1} (1-x^{\frac{3}{2}})^{\frac{2}{3}})\sqrt{1+(-\frac{\sqrt{x}}{\sqrt[3]{1-x^\frac{3}{2}}})^2}.$$

BvU
Wi_N said:
thanks, it was for area, not volume. and it was not correct since its int fx * sqrt(1+(f'x)^2) still I am not getting anything i can solve..

$$S=2\pi\int_{0}^{1} (1-x^{\frac{3}{2}})^{\frac{2}{3}})\sqrt{1+(-\frac{\sqrt{x}}{\sqrt[3]{1-x^\frac{3}{2}}})^2}.$$
Wolfram integrator time?

WWGD said:
Wolfram integrator time?

wolfram says solutions don't exist.

Wi_N said:
wolfram says solutions don't exist.
Does it mean the function is not integrable or it does not have a " nice form".

WWGD said:
Does it mean the function is not integrable or it does not have a " nice form".

true. but I am doing somethng wrong. maybe i can't convert $$x^{\frac{3}{2}} + y^{\frac{3}{2}}=1$$ to $$y(x)=(1-x^{\frac{3}{2}})^{\frac{2}{3}}.$$ and solve it using single integral..

im thinking of $$(1-x^{\frac{3}{2}})=t$$ ...but i can't wiggle anything out of that..

i think $$t=x/(1-x^{\frac{3}{2}})$$ will be fruitful. checking that out now.

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is there any other method of solving this?

GOT IT. $$t=(1-x^{\frac{3}{2}})^{2/3}$$

edit: i think.

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When wolfram alpha determines the integral, it expresses the answer in terms of gamma functions.

Wi_N
NascentOxygen said:
Wolfram alpha expresses the integral in terms of gamma functions.
this is year freaking 1 mathematics calc 101.

edit: supposedly.

any other method you can think of of how i can solve it. i think i made an error somewhere because that integral is waay too hard.

Do you know the correct answer? W-alpha says the answer is a little over 4¼. The surface area of your half coconut shell shape, is that what you're after?

Wi_N
NascentOxygen said:
Do you know the correct answer? W-alpha says the answer is a little over 4¼.

no i don't =(. i feel like there is a short cut somewhere? some other method of solving the core problem?

edit what dd you type into wolfram? cause it just tells me no solution exists.

what about parameter change to trigs? double integrals? anything like that?

edit: is there a standard primitive function for solving xsqrt(1+(x')^2)

NascentOxygen said:
Do you know the correct answer? W-alpha says the answer is a little over 4¼. The surface area of your half coconut shell shape, is that what you're after?

yes. it wants the area that rotates the x-axis. think first year calc 101. it shouldn't be that hard.

Did you arrive at the right formula? What's all that under the square root sign?? I don't have that.

And if you are still able to edit, can you go back to your first post and correct it—to calculating surface area of a solid, not volume. It's not going to be a double integral, either.

NascentOxygen said:
Did you arrive at the right formula? What's all that under the square root sign?? I don't have that.

And if you are still able to edit, can you go back to your first post and correct it—to calculating surface area of a solid, not volume. It's not a double integral, either.

I can't edit it. Here is the full question:

Calculate the area of the rotationsurface that the curve x^(3/2) + y^(3/2) =1 creates around the x-axis.

Had you specified the problem fully, and sketched your working, this may have gone more smoothly.

Are you slicing the solid vertically, into thin slices each of thickness dx, and writing the equation for that slice's surface area?

NascentOxygen said:
Had you specified the problem fully, and sketched your working, this may have gone more smoothly.

Are you slicing the solid vertically, into thin slices each of thickness dx, and writing the equation for that slice's surface area?

did i mention I am a first semester calc 101 student. I am not doing real analysis here.
how would you solve this question?

well right now i have a integral that is unsolvable..and i haven't heard of any other method of solving it.

You need to show your working, so that others can see what you are doing right, and point out where you are going wrong.

NascentOxygen said:
You need to show your working, so that others can see what you are doing right, and point out where you are going wrong.

BvU
NascentOxygen said:
W-alpha says the answer is a little over 4¼
Funny. I see 5.5 (also when I do it in Excel) ? And Wolfram gives no clue of anything analytic.

I sympathize with Wi_N -- this thing looks, and is, horrible. Perhaps we are overlooking something, as @Wi_N suspects in #21, but I for sure don't see it. All I can think of is the numerical solution, but that's not appropriate for calc 101. So my best guess is teacher (or the exercise composer) made a mistake somewhere.
Let us know how this goes !

##\ ##

Wi_N
BvU said:
Funny. I see 5.5 (also when I do it in Excel) ? And Wolfram gives no clue of anything analytic.

I sympathize with Wi_N -- this thing looks, and is, horrible. Perhaps we are overlooking something, as @Wi_N suspects in #21, but I for sure don't see it. All I can think of is the numerical solution, but that's not appropriate for calc 101. So my best guess is teacher (or the exercise composer) made a mistake somewhere.
Let us know how this goes !

##\ ##
thank you so much for your help. i get the answer tomorrow on friday. i will make sure to update you in this thread.

BvU and WWGD
Update:
Showed my professor the problem he stared at it for a few minutes and said he would get back to me. The questions are automated so there is a chance i received a question that is not solvable. My professor is world famous (he has a wikipedia page) so I don't think its solvable.

WWGD and BvU
Was this problem solved finally? or is it not solvable?

## 1. How do you calculate the surface area of a solid of rotation?

The surface area of a solid of rotation can be calculated using the formula 2π∫abf(x)√(1+(f'(x))^2)dx, where a and b are the limits of integration and f(x) is the function that defines the shape of the solid.

## 2. What is a solid of rotation?

A solid of rotation is a three-dimensional object formed by rotating a two-dimensional shape around an axis. Examples include cylinders, cones, and spheres.

## 3. What is the difference between surface area and volume?

Surface area is the measure of the total area of the outer surface of a three-dimensional object, while volume is the measure of the space occupied by the object.

## 4. Can the surface area of a solid of rotation be calculated using basic geometry?

No, the surface area of a solid of rotation cannot be calculated using basic geometry formulas. It requires the use of calculus and integration to find the exact surface area.

## 5. What are some real-life applications of calculating the surface area of a solid of rotation?

Calculating the surface area of a solid of rotation is important in fields such as engineering, architecture, and manufacturing. It is used to determine the amount of material needed to create a specific shape or to estimate the strength and stability of a structure.

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