# Calculating the surface area of a solid of rotation

## Homework Statement:

Calculate surface area of the solid when a curve is rotated around x axis

## 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?

Related Calculus and Beyond Homework Help News on Phys.org
BvU
Homework Helper
What is it you want to calculate ? Is there a complete problem statement ? Volume or area ?

• Wi_N
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).

Last edited:
BvU
Homework Helper
By all means !
Did you have bounds for the integration in mind ?

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?

BvU
Homework Helper
Yes

• Wi_N
Mark44
Mentor
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
BvU
Homework Helper
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 ...

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

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..

Last edited by a moderator:
BvU
Homework Helper
$$S=2\pi\int_{0}^{1} (1-x^{\frac{3}{2}})^{\frac{2}{3}})\sqrt{1-(-\frac{\sqrt{x}}{\sqrt{1-x^\frac{3}{2}}})^2}.$$

(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 $y=\Bigl (1-x^{3\over 2} \Bigr)^{2\over 3}$ . $$S=2\pi\int_{0}^{1} (1-x^{\frac{3}{2}})^{\frac{2}{3}})\sqrt{1-(-\frac{\sqrt{x}}{\sqrt{1-x^\frac{3}{2}}})^2}.$$ (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 im not getting anything i can solve..

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

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

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

Wolfram integrator time?
wolfram says solutions don't exist.

WWGD
Gold Member
wolfram says solutions don't exist.
Does it mean the function is not integrable or it does not have a " nice form".

Does it mean the function is not integrable or it does not have a " nice form".
true. but im 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.

Last edited:
is there any other method of solving this?

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

edit: i think.

Last edited by a moderator:
NascentOxygen
Staff Emeritus
When wolfram alpha determines the integral, it expresses the answer in terms of gamma functions.

• Wi_N
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.

NascentOxygen
Staff Emeritus
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
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)

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.

NascentOxygen
Staff Emeritus
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.

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.

NascentOxygen
Staff Emeritus