Results 1 to 13 of 13

Math Help - Confirmation of Thoughts/Results on a Few Double Integration Problems

  1. #1
    Junior Member
    Joined
    Jul 2010
    Posts
    41

    Confirmation of Thoughts/Results on a Few Double Integration Problems

    Hello there,

    I'm tutoring somebody at the moment and I'm having some trouble with a few double integral questions that have cropped up in their past papers. I'm not sure whether or not there are errors in the questions that make them impossible to tackle or whether I'm just over thinking things, so I'm looking for some confirmation. I'll give you the question, my final answer and explain my issue, and hopefully you guys can clear things up.

    1) Sketch the domain of integration and evaluate the integral

     I = \int_1^3 \int_{2y}^{0} (yx^2+xy^2) \, dx \, dy

    For this I can evaluate the integral and get an answer of I =  -\frac{3888}{15} , but I can't find a way to sketch the region.

    I've drawn a graph with the lines  y = 1 \,\,\, y = 3 \,\,\, x = 0 \,\,\, x = 2y , but there is no region enclosed by these curves, so how can the region be sketched/specified? Also, if no such region exists, how is it possible to evaluate the integral?

    2) Evaluate  \displaystyle I = \int_{1}^{2x} \int_0^2 \left(yx^3 + xy^2\right) \, dx \, dy

    Here I am able to evaluate the integral and arrive at the result:

     I = \frac{16}{3}x^3 + 8x^2 - \frac{8}{3}.

    However, I'm slightly concerned that I'm getting a function of x as a result rather than a numerical result... is this correct?

    Either I'm correct and I'm over-thinking it, or it is an erroneous question (any double integrals I've seen in the past have the variable limits on the inner integral and the numerical limits on the outer integral, hence giving a numerical result), OR there is another way to deal with double integrals with variable limits on the outer integral that I'm not yet aware of. Which is it?

    3) The height ( z) of the ceiling of an art gallery is defined by the surface:

     z = 5+xy

    The gallery wall is circular (in plan) and is defined by the equation:

    x^2 + y^2 = 36

    For  y > 0 , derive an integral expression for the area of the wall between the limits  0 \leq x \leq 5 .

    For this one I'm not sure where to get started at all. It might not even be a double integral, but some sort of line integral or something? Anybody have any hints?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    GJA
    GJA is offline
    Member
    Joined
    Jul 2012
    From
    USA
    Posts
    109
    Thanks
    29

    Re: Confirmation of Thoughts/Results on a Few Double Integration Problems

    Hi, Phugoid. I'm a little confused about the region for problem #1 as well.

    For problem #3 do they want the surface area of the ceiling along with the outer walls?

    To find the surface area of the ceiling you can use the formula

    \int\int\sqrt{1+f_{x}^{2}+f_{y}^{2}}dA,

    where f(x,y)=5+xy. See Area - Wikipedia, the free encyclopedia about three quarters of the way down for this formula.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2010
    Posts
    41

    Re: Confirmation of Thoughts/Results on a Few Double Integration Problems

    Quote Originally Posted by GJA View Post
    Hi, Phugoid. I'm a little confused about the region for problem #1 as well.
    Doesn't make any sense at all, does it?

    For problem #3 do they want the surface area of the ceiling along with the outer walls?

    To find the surface area of the ceiling you can use the formula

    \int\int\sqrt{1+f_{x}^{2}+f_{y}^{2}}dA,

    where f(x,y)=5+xy. See Area - Wikipedia, the free encyclopedia about three quarters of the way down for this formula.
    It is my interpretation of #3 that they want the area of the wall alone, given that the height of the wall is dependent on the value of z along the curve y = \sqrt{36-x^2} for  0 \leq x \leq 5 .

    Basically an "area under a curve" problem except that the area of projected onto the inside of a circle rather than a flat surface. If that is the case, I would probably proceed as follows:

    I = \int_S z \, ds\\ \newline=\int_S (5+xy) \, ds\\ \newline =\int_S (5+xy) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\\ \newline=\int_{x_1}^{x_2} (5+xy)\sqrt{1 + \left(\frac{x^2}{36 - x^2}\right)} \, dx\\ \newline =\int_{0}^{5} (5+xy)\sqrt{1 + \left(\frac{x^2}{36 - x^2}\right)} \, dx \\ \newline =\int_{0}^{5} (5+x\sqrt{36 - x^2})\sqrt{1 + \left(\frac{x^2}{36 - x^2}\right)} \, dx

    Anybody concur?

    What are your thoughts on #2?
    Last edited by Phugoid; August 5th 2012 at 03:33 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121

    Re: Confirmation of Thoughts/Results on a Few Double Integration Problems

    Quote Originally Posted by Phugoid View Post
    Hello there,

    I'm tutoring somebody at the moment and I'm having some trouble with a few double integral questions that have cropped up in their past papers. I'm not sure whether or not there are errors in the questions that make them impossible to tackle or whether I'm just over thinking things, so I'm looking for some confirmation. I'll give you the question, my final answer and explain my issue, and hopefully you guys can clear things up.

    1) Sketch the domain of integration and evaluate the integral

     I = \int_1^3 \int_{2y}^{0} (yx^2+xy^2) \, dx \, dy

    For this I can evaluate the integral and get an answer of I =  -\frac{3888}{15} , but I can't find a way to sketch the region.

    I've drawn a graph with the lines  y = 1 \,\,\, y = 3 \,\,\, x = 0 \,\,\, x = 2y , but there is no region enclosed by these curves, so how can the region be sketched/specified? Also, if no such region exists, how is it possible to evaluate the integral?
    I don't know why you say that. y= 1 and y= 3 are horizontal lines. x= 0 crosses those lines at (0,1) and (0,3), of course. And x= 2y crosses y= 1 at x= 2 and y= 3 at x= 6. That is, the region is the trapezoid with vertices at (0,1), (2,1), (6, 3), and (0,3). Looks straightforward to me.

    2) Evaluate  \displaystyle I = \int_{1}^{2x} \int_0^2 \left(yx^3 + xy^2\right) \, dx \, dy

    Here I am able to evaluate the integral and arrive at the result:

     I = \frac{16}{3}x^3 + 8x^2 - \frac{8}{3}.

    However, I'm slightly concerned that I'm getting a function of x as a result rather than a numerical result... is this correct?

    Either I'm correct and I'm over-thinking it, or it is an erroneous question (any double integrals I've seen in the past have the variable limits on the inner integral and the numerical limits on the outer integral, hence giving a numerical result), OR there is another way to deal with double integrals with variable limits on the outer integral that I'm not yet aware of. Which is it?
    Technically, it is not an error. The first "x", as a variable of integration, is a "dummy" variable and there is no "x" in the problem after you have done the first integral so you can legally use "x" as a limit of integration in the second integral. Of course, it would be easier to understand if we used a different dummy variable- say \int_0^{2x}\int_0^2 yt^3+ ty^2 dt dy.

    3) The height ( z) of the ceiling of an art gallery is defined by the surface:

     z = 5+xy

    The gallery wall is circular (in plan) and is defined by the equation:

    x^2 + y^2 = 36

    For  y > 0 , derive an integral expression for the area of the wall between the limits  0 \leq x \leq 5 .

    For this one I'm not sure where to get started at all. It might not even be a double integral, but some sort of line integral or something? Anybody have any hints?
    No, its a pretty standard integral over an area. The only result of the last limits is that instead of integrating, with respect to x, from -6 to 6 (covering the entire circle), you integrate from 0 to 5. For each x, of course, y= \pm\sqrt{36- x^2}. The integral is
    \int_{x=0}^5\int_{y=-\sqrt{36- x^2}}^{\sqrt{36- x^2}} 5+ xy dydx
    Follow Math Help Forum on Facebook and Google+

  5. #5
    GJA
    GJA is offline
    Member
    Joined
    Jul 2012
    From
    USA
    Posts
    109
    Thanks
    29

    Re: Confirmation of Thoughts/Results on a Few Double Integration Problems

    If it's just the wall under the circular arc, you can do a line integral by parameterizing the arc of the circle you're integrating over.

    I'm a little confused on #2 as well. Maybe I'm forgetting something, or maybe they do want your answer in terms of x, but I can't say I recall an example I've worked like that recently.

    Update: For some reason the site is not letting me post new things, so I am forced to respond to the threads by updating my previous post.

    First, I agree with your most recent post, Phugoid. It seems to me that to get the inequalities on x are not correct.

    Second, I don't think you want to do the area of the wall problem as a double integral; that would give you the volume of the room. I think doing a line integral by parameterizing the arc you're interested in is the way to go.
    Last edited by GJA; August 5th 2012 at 03:43 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jul 2010
    Posts
    41

    Re: Confirmation of Thoughts/Results on a Few Double Integration Problems

    Quote Originally Posted by HallsofIvy View Post
    I don't know why you say that. y= 1 and y= 3 are horizontal lines. x= 0 crosses those lines at (0,1) and (0,3), of course. And x= 2y crosses y= 1 at x= 2 and y= 3 at x= 6. That is, the region is the trapezoid with vertices at (0,1), (2,1), (6, 3), and (0,3). Looks straightforward to me.
    I agree that there is a region bounded by those lines, but because 2y \leq x \leq 0 I thought it should be the case that the lower bound of the region should by the line  x = 2y rather than the line  x = 0 and vice versa.

    For me, the region you describe is  1 \leq y \leq 3 \,\,\,\, 0 \leq x \leq 2y and not  1 \leq y \leq 3 \,\,\,\, 2y \leq x \leq 0

    Technically, it is not an error. The first "x", as a variable of integration, is a "dummy" variable and there is no "x" in the problem after you have done the first integral so you can legally use "x" as a limit of integration in the second integral. Of course, it would be easier to understand if we used a different dummy variable- say \int_0^{2x}\int_0^2 yt^3+ ty^2 dt dy.
    I understand, but I was just concerned by the fact that the result of the whole integral is a function of x rather than a numerical value. I've never seen a double integral like that before - usually the "variable" limits are on the inner integral, and the numeral limits ar eon the outer integral, hence always resulting in a numerical result in the evaluation of the integral.

    No, its a pretty standard integral over an area. The only result of the last limits is that instead of integrating, with respect to x, from -6 to 6 (covering the entire circle), you integrate from 0 to 5. For each x, of course, y= \pm\sqrt{36- x^2}. The integral is
    \int_{x=0}^5\int_{y=-\sqrt{36- x^2}}^{\sqrt{36- x^2}} 5+ xy dydx
    Not sure where you got that double integral from. It states that you're supposed to evaluate it for y > 0 so how does the lower limit of your inner integral come into place?

    Also do you agree with my working for that question in post #3?
    Last edited by Phugoid; August 5th 2012 at 03:42 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jul 2010
    Posts
    41

    Re: Confirmation of Thoughts/Results on a Few Double Integration Problems

    Quote Originally Posted by HallsofIvy View Post
    .
    BUMP!

    Anyone?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,969
    Thanks
    1010

    Re: Confirmation of Thoughts/Results on a Few Double Integration Problems

    For the first one, the property \displaystyle \begin{align*} \int_b^a{f(x)\,dx} = -\int_a^b{f(x)\,dx} \end{align*} should help you...
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jul 2010
    Posts
    41

    Re: Confirmation of Thoughts/Results on a Few Double Integration Problems

    Quote Originally Posted by Prove It View Post
    For the first one, the property \displaystyle \begin{align*} \int_b^a{f(x)\,dx} = -\int_a^b{f(x)\,dx} \end{align*} should help you...
    I agree. I'm satisfied with that one now.

    For #2 I'd just like for somebody to confirm that it's OKAY to get a function of x for an answer to the integral - or, if not, then how do you deal with that integral to GET a numerical answer. And for #3 could anybody confirm that they agree with my working?

    I tried a simpler test case for #3 - i.e. if the ceiling was just a constant 5, say, and we were interested in the area of the wall for 0 <= x <= 6, then my integral comes out to be  15 \pi , and it is easy to verify that this would be the area of a wall that's composed of a quarter circle of radius 6 and height 5. So on that basis, I'm fairly confident of my result - if anybody could corroborate, I would be happy!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,969
    Thanks
    1010

    Re: Confirmation of Thoughts/Results on a Few Double Integration Problems

    Quote Originally Posted by Phugoid View Post
    I agree. I'm satisfied with that one now.

    For #2 I'd just like for somebody to confirm that it's OKAY to get a function of x for an answer to the integral - or, if not, then how do you deal with that integral to GET a numerical answer. And for #3 could anybody confirm that they agree with my working?

    I tried a simpler test case for #3 - i.e. if the ceiling was just a constant 5, say, and we were interested in the area of the wall for 0 <= x <= 6, then my integral comes out to be  15 \pi , and it is easy to verify that this would be the area of a wall that's composed of a quarter circle of radius 6 and height 5. So on that basis, I'm fairly confident of my result - if anybody could corroborate, I would be happy!
    I expect with 2, you need to switch the order of integration (note you won't need to change your bounds, check the region with a sketch if you must).
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Jul 2010
    Posts
    41

    Re: Confirmation of Thoughts/Results on a Few Double Integration Problems

    Quote Originally Posted by Prove It View Post
    I expect with 2, you need to switch the order of integration (note you won't need to change your bounds, check the region with a sketch if you must).
    So you're saying I can go:

     \displaystyle I = \int_{1}^{2x} \int_0^2 \left(yx^3 + xy^2\right) \, dx \, dy =  \int_0^2 \int_{1}^{2x} \left(yx^3 + xy^2\right) \, dy \, dx

    Can you explain why I don't need to change the bounds in this case? I thought it was only for the integration of a Rectangle that it would not be necessary to change the bounds in a change of order of integration?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,969
    Thanks
    1010

    Re: Confirmation of Thoughts/Results on a Few Double Integration Problems

    Quote Originally Posted by Phugoid View Post
    So you're saying I can go:

     \displaystyle I = \int_{1}^{2x} \int_0^2 \left(yx^3 + xy^2\right) \, dx \, dy =  \int_0^2 \int_{1}^{2x} \left(yx^3 + xy^2\right) \, dy \, dx

    Can you explain why I don't need to change the bounds in this case? I thought it was only for the integration of a Rectangle that it would not be necessary to change the bounds in a change of order of integration?
    I mean because it appears that your bounds are ALREADY written as though you're summing up horizontal strips. If you had to change from horizontal to vertical, then you would have to change the bounds...
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Jul 2010
    Posts
    41

    Re: Confirmation of Thoughts/Results on a Few Double Integration Problems

    Quote Originally Posted by Prove It View Post
    I mean because it appears that your bounds are ALREADY written as though you're summing up horizontal strips. If you had to change from horizontal to vertical, then you would have to change the bounds...
    Ah, I see what you mean. The integration limits are written as if they're summing up the horizontal strips (i.e.  \,dy \,dx), but for some reason the question has it written  \, dx \,dy instead. So it's just a case of correct that by writing:

     \displaystyle \int_1^{2x} \int_0^2 (yx^3+xy^2) \, dy \, dx

    I'm not really changing the order of integration, I'm just correcting the initial expression because it was written inappropriately.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  2. Replies: 1
    Last Post: April 22nd 2010, 05:59 PM
  3. Replies: 1
    Last Post: March 1st 2010, 03:57 AM
  4. Integration By Parts Confirmation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 28th 2009, 03:09 AM
  5. Need confirmation of 2 trig problems, please...
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 7th 2009, 11:25 AM

Search Tags


/mathhelpforum @mathhelpforum