Reversing the order of integration in a double integral

[dyn]

Homework Statement

Calculate the double integral of y^2 - 4x over the following region -y^2 < x < y^2 , 0 < y < 1

Relevant Equations

I did the integral by dividing the region into strips parallel to the x-axis and got the answer 2/5 which i know is correct but i am trying to reverse the order of integration but i can't get the same answer

Performing the x-integration first the limit are x=y² and x= -y² and then the y limits are 0 to 1. This gives the final answer 2/5

But i am getting confused when trying to reverse the order of integration. My attempt is that i have to divide the region in 2 equal halfs and then double my answer at the end. I used y limits from √x to 1 and x limits from 0 to 1 but i am ending up with the answer zero ! Where am i going wrong with my limits here ?
Thanks

 

[anuttarasammyak]

The integration is
$$\int_S (y^2-4x) \ dS$$
Area S is two-side-curved triangle with vertex (0,0),(1,-1),(1,1)

My attempt is that i have to divide the region in 2 equal halfs and then double my answer at the end.

That' good for integrand y^2 but not for -4x because of antisymmetry it cancels by right plus and left minus of y-axis
$$\int_S (-4x) \ dS=0$$
You should omit it from integration calculation in right half.

 

[dyn]

The answer of the total integral is 2/5 which i got my dividing the region into horizontal strips. If i divide the region into 2 equal halfs along the line x=0 and perform the 2 integrals using horizontal strips i get I_R = 0 and I_L = 2/5 which when added together give my total answer of 2/5.

If i now use vertical strips for both halfs , i can get I_R = 0 as it should but i cannot get 2/5 for I_L. I am struggling with the limits because they involve square roots of negative numbers. Is this an example of where reversing the order of integration is impossible to perform or extremely difficult ? Or am i just missing something when trying to perform the integral on the LHS using vertical strips ?

 

[anuttarasammyak]

$$I=\int (y^2-4x) dS = \int y^2\ dS -4 \int x\ dS$$
By vertical strips for halfs
$$I=\int y^2\ dS_R+ \int y^2\ dS_L -4 \int x\ dS_R-4 \int x\ dS_L$$
$$=2\int y^2\ dS_R$$
because
$$\int y^2\ dS_R= \int y^2\ dS_L$$
$$\int x\ dS_R=- \int x\ dS_L$$
You said right half integration of y^2-4x is zero so I assume
$$I=1/5 + 1/5 - 1/5 + 1/5$$
Confirm this by integration of each term if you please.

 

[dyn]

To perform the integral over the RH region using vertical strips i integrate (y²-4x) wrt y from √x to 1 ; i then integrate 1/3 - 2x +(5/3)x^3/2 wrt x from 0 to 1. This gives a final answer of 0 which is the same answer as expected if i integrated using horizontal strips.
The problem comes when i try to integrate y²=4x over the LH side using vertical strips

 

[anuttarasammyak]

The problem comes when i try to integrate y2=4x over the LH side using vertical strips

$$\int^0_{-1} dx \int_{\sqrt{-x}}^1 dy (y^2-4x)=\int^0_{-1} dx\ [y^3/3-4xy]_{y=\sqrt{-x}}^{y=1}$$
RH side
$$\int^1_0 dx \int_{\sqrt{x}}^1 dy (y^2-4x)=\int^1_0 dx\ [y^3/3-4xy]_{y=\sqrt{x}}^{y=1}$$

 

[dyn]

i apologise ; i was writing this late at night. The function to be integrated is y²- 2x
It's late again so i will look at this again when i wake

 

[anuttarasammyak]

Do not worry. It does not matter on the result of integration.

 

[etotheipi]

N.B. you can also do$$\begin{align*}
\int_S (y^2 - 4x) dy dx &= 2\int_{S_R} y^2 dy dx\\

&= 2\int_0^1 \int_0^{y^2} y^2 dx dy\\

&= 2\int_0^1 y^4 dy\\

&= \frac{2}{5}

\end{align*}
$$which also agrees with @anuttarasammyak's solution.

 

[dyn]

$$\int^0_{-1} dx \int_{\sqrt{-x}}^1 dy (y^2-4x)=\int^0_{-1} dx\ [y^3/3-4xy]_{y=\sqrt{-x}}^{y=1}$$

I am still not getting the correct answer for integral using vertical strips over the LHS. The function to be integrated is actually y² -2x
Using the limits you have given i get
1/3 - 2x -(5/3)(-x)^5/2 to be integrated from -1 to 0 giving me a final answer of 2 which is still wrong

 

[anuttarasammyak]

Changing the coefficient to 2 and introducing -x=t
$$I_L=\int_0^1 dt [y^3/3 +2ty]_{y=\sqrt t}^{y=1}=I_1+I_2$$
corresponding to
$$I_R=\int_0^1 dx [y^3/3 -2xy]_{y=\sqrt x}^{y=1}=I_1-I_2$$
where
$$I_1=\int_0^1 dt [y^3/3]_{y=\sqrt t}^{y=1}=1/5$$
$$I_2=\int_0^1 dt [2ty]_{y=\sqrt t}^{y=1}=1/5$$
though we do not have to calculate $I_2$ to get the answer.

 

[dyn]

The answer should be 2/5 and the function to be integrated is y² -2x.

Integrating this wrt y gives y³/3 - 2xy

Inserting the limits of y from√(-x) to1 gives the following

1/3 - 2x - [ (-x)^3/2/3 - 2x(-x)^1/2] = 1/3 - 2x - (7/3)(-x)^3/2

Integrating this wrt x gives x/3 - x² - (14/15)(-x)^5/2

Inserting the limits of x from -1 to 0 give the final answer

(1/3) + 1 + (14/15) = 34/15

I have spent hours on this problem in total and i can't get the answer 2/5 by using vertical strips ! It is easy using horizontal strips but i wanted to check that i could do it using vertical strips but i can't !

 

[Charles Link]

I think the substitution mentioned in post 11 is the best way to proceed. Otherwise I think the $-x$ to the fractional power is causing problems that aren't readily resolved.
Edit: I think I see the problem though: In the integration, in going from $3/2$ to $5/2$ in the exponent, you introduce another minus sign, which alternatively, you can apply the chain rule and check your integration by taking derivatives, and I think you are off by a minus sign on that term.

 

[anuttarasammyak]

1/3 - 2x - [ (-x)3/2/3 - 2x(-x)1/2] = 1/3 - 2x - (7/3)(-x)3/2
Integrating this wrt x gives x/3 - x2 - (14/15)(-x)5/2

How did you integrate the RHS last term,
$$-\frac{7}{3}\int_{-1}^0 (-x)^{3/2} dx$$?

 

[Charles Link]

See also my edit in post 13.

 

[vela]

Integrating this wrt x gives x/3 - x² - (14/15)(-x)^5/2

Inserting the limits of x from -1 to 0 give the final answer

(1/3) + 1 + (14/15) = 34/15

Sign errors when you plugged in -1.

 

[Charles Link]

Observe: $dy/dx=(-7/3)(-x)^{3/2}$.
$y=+(14/15)(-x)^{5/2}$. Take the derivative with the chain rule, and you will see I have the sign right. The OP has it incorrect.

 

[dyn]

Thank you everyone so much. It was driving me nuts ! Yes , it was that negative sign when integrating the (-x). I am so relieved. Thank you all again !