Calculating Length and Area of y=3sin(x+x^2)

[jpd5184]

Homework Statement

find the length of y=3sin(x+x^2)
find its length and area

Homework Equations

length: integral of square root of 1+ (dy/dx)2 dx
area: integral 2pix square root of 1 + (dy/dx)^2 dx

The Attempt at a Solution

dy/dx = 3cos(x+x^2)(2x)

 

[SammyS]

No end points are specified for the arc length.

There is no enclosed area defined.

 

[jpd5184]

sorry about that:

0 < x < pi/2

both are greater than or equal to and less than or equal to.

 

[JThompson]

Homework Statement

The Attempt at a Solution

dy/dx = 3cos(x+x^2)(2x)

dy/dx = 3cos(x+x^2)(2x+1), since the argument of sine is x+x^2.

 

[SammyS]

For area, do you mean the surface area produced by revolving the arc about the y-axis?

That's what the 2πx suggests.

 

[TylerH]

You need to do:
\int_{0}^{2\pi}\sqrt{1+\left( \frac{dy}{dx} \right)^2}dx to get the arc length, and
\int_{0}^{2\pi}ydx to get the area,
where y=3sin(x+x^2).

 

[jpd5184]

yes around the y-axis.

i need help integrating dy/dx

do i use u substitution or can i just integrate right away

 

[jpd5184]

You need to do:
\int_{0}^{2\pi}\sqrt{1+\left( \frac{dy}{dx} \right)^2}dx to get the arc length, and
\int_{0}^{2\pi}ydx to get the area,
where y=3sin(x+x^2).

dy/dx = 3cos(x+x^2)(2x+1)

im not so sure how to to the integral though.

 

[SammyS]

dy/dx = 3cos(x+x^2)(2x+1)

I'm not so sure how to to the integral though.

JThompson showed you what dydx is.

Plug it in & give it a whirl.

 

[jpd5184]

so its the integral of the square root of 1+ (3cos(x+x^2)(2x+1))^2

integral of cos(x) is sin(x)

would i use u substitution.

then it would be 1+ u^3/3

 

[TylerH]

I don't know. But, I bet Wolfram Alpha does.

 

[Dick]

I don't know. But, I bet Wolfram Alpha does.

And I'd bet nobody can integrate stuff like that in elementary analytic form. Not even WA. Is this a question where you are supposed to just give a numerical approximation?

 

[jpd5184]

what do you mean it can be integrated. why not. its a question i was asked in class so i would think it can be solved.

 

[Dick]

what do you mean it can be integrated. why not. its a question i was asked in class so i would think it can be solved.

I mean that there are a lot of simple looking functions that don't have a simple form for the integral. You can write down an integral form for the length and area, which may be all you are expected to do. You could also try and numerically approximate them. But you probably can't simplify the integral expressions much.