Calculate the following double integral

In summary, the student attempted to find the double integral for the homework equation, but found it difficult. After trying to find an interval to fit the curve, the student realized that an enclosed D was needed, and after finding the interval, found the two integrals to be very easy.
  • #1
Hannisch
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0

Homework Statement


Calculate the double integral:

[tex]\iint\limits_D x^{5}y^{6}dxdy[/tex]
where D = {(x,y): x9 ≤ y ≤ x1/9}

Homework Equations





The Attempt at a Solution



I didn't think this problem would be too hard, but it seems I'm really not good with double integrals.

Anyway, I first tried to find the right interval. The curve y=x9 is below y=x1/9 in two intervals, first of all 0 ≤ x ≤ 1 and -∞ ≤ x ≤ -1. I was wondering about this for some time, but then I thought that I probably needed an enclosed D, so I discarded the second interval.

When x=0, y=0, and when x=1, y=1 in both curves.

[tex]\iint\limits_D x^{5}y^{6}dxdy = \int ^{1}_{0} x^5 dx \int ^{1}_{0} y^6 dy[/tex]

I tried that, because that was something they explained to us during the lecture and well.. it seemed to apply. And those two integrals are really very easy:

[tex]\int ^{1}_{0} x^5 dx \int ^{1}_{0} y^6 dy = \left[ \frac{1}{6} x^6 \right] ^{1}_{0} \left[ \frac{1}{7} y^7 \right]^{1}_{0} = ( \frac{1}{6})( \frac{1}{7}) = \frac{1}{42}[/tex]

And this is wrong.

Help?
 
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  • #2
The two curves intersect at (-1, -1), (0, 0), and (1, 1). When 0 <= x <= 1, x^(1/9) >= x^9, and when -1 <= x <= 0, x^9 >= x^(1/9). So that means you'll need to split your double integral into a pair of iterated integrals, one for each of the two intervals.
Hope that helps.
 
  • #3
You must integrate over y first because the limits of the y integral contain x values. After the y integral, you do the x integral. You should end up with an expression with only x's.
 
  • #4
Also, I don't think there's a justification for splitting up the iterated integrals as you did. Flatmaster's advice of integrating with respect to y first, and then with respect to x, is good.
 
  • #5
Oh thank you! Talk about change in thinking.

And anyway, I didn't use that second interval, because it didn't fit with the area that I was given, since it stated that x^(1/9) needed to be >= x^9. It gave me the right answer, so I'm very happy now!
 

What is a double integral?

A double integral is a type of mathematical operation that calculates the volume under a surface in a 3D space. It involves taking the area between two curves and multiplying it by the width of the space.

What is the difference between a single and double integral?

A single integral calculates the area under a curve in a 2D space, while a double integral calculates the volume under a surface in a 3D space. In other words, a single integral is a 2D operation, while a double integral is a 3D operation.

What is the purpose of calculating a double integral?

The purpose of calculating a double integral is to find the volume under a surface in a 3D space. This can be useful in many different fields, such as physics, engineering, and economics.

What are the steps involved in calculating a double integral?

The steps involved in calculating a double integral include: defining the limits of integration, setting up the integral using the appropriate formula, evaluating the integral, and interpreting the result in the context of the problem.

What are some real-world applications of double integrals?

Double integrals have many real-world applications, such as calculating the mass of an object with varying density, finding the center of mass of an irregularly shaped object, and determining the work done by a varying force over a given distance.

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