Double integral by changing order and limits

[s_gunn]

Homework Statement

Evaluate the following double integral by changing the order of integration:

∫(lower 0 and upper 1)∫ (lower √x and upper 1) sin(((y^3)+1)/2) dydx

2. Homework Equations

In case it's not clear from above! y is between √x and 1, and x is between 0 and 1.

The Attempt at a Solution

I can do the integration parts of this but have no idea if I have worked out the new limits right!? Is it x between 0 and y^2 and y between 0 and 1?

Any advice on how to input equations into my posts would be helpful too!?

 

[quantumdude]

Is it x between 0 and y^2 and y between 0 and 1?

Yes, that's right.

Any advice on how to input equations into my posts would be helpful too!?

Check out the following thread:

https://www.physicsforums.com/showthread.php?t=8997

Any time you see an equation formatted in LaTeX you can click on the image and the code will pop up (just make sure your settings allow pop ups!)

 

[s_gunn]

thanks for the fast response! I was just confused by my answer which was 0.666565 (6dp) as normally our problems nearly always end up as obvious fractions.

I obviously rounded up to 2/3 though!

I'll try and input my working out by reading that post if you or anyone else can check it for me!

 

[s_gunn]

$$\int_{0}^{1} \int_{0}^{y^2} \sin{\frac{y^3+1}{2} dxdy$$
$$=\int_{0}^{1} [x\sin{\frac{y^3+1}{2}]^y^2_0 dy$$
$$=\int_{0}^{1} [y^2\sin{\frac{y^3+1}{2}] dy$$

then by inspection:

$$=\frac{2}{3}\int_{0}^{1}\frac{d}{dy}(-cos{\frac{y^3+1}{2}}) dy$$
$$=[\frac{2}{3}(-cos{\frac{y^3+1}{2}})]^1_0$$


$$= 0.666565 \approx\frac{2}{3}$$

Does this seem right?!

I feel so proud inputing all that!

 

[quantumdude]

Everything looks right...except the answer. I think you're having calculator difficulties because when I do the integral I get 0.22485. Maple agrees with me.

 

[s_gunn]

I've had 3 different answers from my calculator so far and non have been 0.22485! I think I'm going crazy!

 

[s_gunn]

I've been going over this again and still can't come up with the answer that maple (and an online definite integral calculator!) gives of 0.22485. I've tried having my calculator in radians (=0.306465) in degrees (=-1.10537 * 10^-4)
Can anyone see what I'm doing wrong - I've never had a problem like this before when my working is right but the numerical answer just won't come out right!

 

[lanedance]

Hi S_gunn

So your result is:
$$=[\frac{2}{3}(-cos{\frac{y^3+1}{2}})]^1_0$$

putting in the limts
$$=\frac{2}{3}(-cos{1}+cos{\frac{1}{2}})]$$
as cos(0) = 1 and cos(pi/2) = 0 decreasing monotonically, so this should be close to the numebr quoted previously. I would always assume radians unless otherwise stated, as this is the natural input by which a trigonometric function is defined.

Is this what you're inputting? i don't have a calculator handy...

 

[s_gunn]

Thanks for all your help everyone! I made the stupidest mistake possible and worked out that

$$\frac{y^3+1}{2}$$ was 0 when y was 0!

Why is it always something so simple that gets you!??

Well, it's all sorted now but I'm sure I'll be back sometime soon!