Use the given transformation to evaluate the integral

In summary, the given transformation x=u/v and y=v simplifies the given region in the first quadrant bounded by y=x/2, y=2x/3, xy=1/2, and xy=3/2. The resulting double integral is 10∫(u from 1/2 to 3/2) ∫(v from √(u/2) to √(3u/2)) (u/v)dvdu. After performing the integration, the final answer is 10ln(√3).
  • #1
ptguard1
13
0
∫∫10xy(dA), where R is the region in the first quadrant bounded by the lines y=x/2 and y=2x/3 and by the hyperbolas xy=1/2 and xy=3/2

The transformations given in the problem (these cannot be altered): x=u/v and y=v
Relevant equations:

The Jacobian - ∂(x,y)/∂(u,v)
The attempt at a solution:

y=(3/2)x: 2v^2=3u
y=x/2: 2v^2=u
xy=1/2: u=1/2
xy=3/2: u=3/2

After making the transformations, I get the following double integral:

10∫(u from 1/2 to 3/2)∫(v from √(u/2) to √(3u/2)) (u/v)dvdu

I feel like my transformations are suppose to result in basic bounds without variables, so I think I am doing this problem incorrectly and can't figure out any other way to go about it.
 
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  • #2
ptguard1 said:
∫∫10xy(dA), where R is the region in the first quadrant bounded by the lines y=x/2 and y=2x/3 and by the hyperbolas xy=1/2 and xy=3/2

The transformations given in the problem (these cannot be altered): x=u/v and y=v

Relevant equations:

The Jacobian - ∂(x,y)/∂(u,v)

The attempt at a solution:

y=(3/2)x: 2v^2=3u
y=x/2: 2v^2=u
xy=1/2: u=1/2
xy=3/2: u=3/2

After making the transformations, I get the following double integral:

10∫(u from 1/2 to 3/2) ∫(v from √(u/2) to √(3u/2)) (u/v)dvdu

I feel like my transformations are suppose to result in basic bounds without variables, so I think I am doing this problem incorrectly and can't figure out any other way to go about it.
That looks good to me.

The transformation does simplify the region, even if it's not rectangular.

Try the integration.
 
  • #3
Wonderful! I performed the integration and got 10ln(√3) and this was correct.
 

FAQ: Use the given transformation to evaluate the integral

What is a transformation in the context of evaluating integrals?

A transformation is a mathematical operation that allows us to change the variables in an integral. This can make the integral easier to solve or evaluate.

How do I use a given transformation to evaluate an integral?

To use a given transformation, you must first identify which variables need to be replaced. Then, you can substitute the new variables into the original integral and solve it using the techniques you have learned in calculus.

Can any integral be evaluated using a transformation?

Not all integrals can be evaluated using a transformation. Some integrals may be too complex or involve functions that cannot be transformed easily. In these cases, other techniques may be needed to evaluate the integral.

How do I choose the best transformation for a given integral?

The best transformation to use will depend on the specific integral and its variables. It may involve trial and error or using a specific formula or substitution rule. It is important to practice and familiarize yourself with different transformations to determine the best one to use.

Are there any common mistakes to avoid when using a transformation to evaluate an integral?

One common mistake is forgetting to substitute the new variables back into the original integral after solving it. It is also important to check for any restrictions on the new variables, such as domain or range, to ensure the integral is still valid. Another mistake is using the wrong transformation or applying it incorrectly, so it is important to double-check your work and practice regularly to avoid these errors.

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