Polar Double Integral: Converting and Evaluating

In summary: I got the correct answer by plugging in the two expressions and getting 10.6 for the rectangular one and 10.8 for the polar one.In summary, the student attempted to solve a homework problem but was not able to get the correct answer.
  • #1
dtl42
119
0

Homework Statement


Given the integral: [tex]\int_1^2\int_{\frac{3}{\sqrt{x}}}^{\sqrt{3}x}{{(x^2+y^2)}^{\frac{3}{2}}}dy \; dx[/tex]. Convert to polar and evaluate.

Homework Equations


[tex]r=\sqrt{(x^2+y^2)}[/tex]


The Attempt at a Solution


Ok, I've gotten bounds on [tex]\theta[/tex], [tex]\frac{\pi}{6} \le \theta \le \frac{\pi}{3}[/tex]. I'm not sure what the bounds for r should be, otherwise I have the integral:
[tex]\int_\frac{\pi}{6}^\frac{\pi}{3}\int_{?}^{?}{r^3}dr \; d\theta[/tex]
 
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  • #2
Do you have that x-y integral right? Did you draw a sketch of the region? At x=1 the lower limit of y is 3 and the upper limit is sqrt(3). At x=2 the lower is 3/sqrt(2) and the upper is 2*sqrt(3). The two limits are equal at theta=pi/3. This sounds really funny. At the very least, I don't believe your theta limits.
 
  • #3
That is the rectangular integral given in the book. I think the region is a quadrilateral-like thing that is enclosed by x=1,2 and y=rt(3)x, x/rt(3). I found the limits for theta by setting tan(theta)=rt(3), 1/rt(3) because then the thetas produce the necessary lines.
 
  • #4
That helps. In the problem statement you wrote 3/sqrt(x), not x/sqrt(3). Yes, the region is a trapezoid (a quadrilateral-like thing). Now I agree with your theta limits. That means your r limits will be a function of theta, right? Find them by intersecting a line at an angle theta through the origin with the two vertical lines in your quadrilateral-like thing.
 
  • #5
Wow, I'm sorry for that typo. That would change things. Would the correct limits for r be [tex]\frac{1}{cos(\theta)}\text{and}\frac{2}{cos(\theta)}[/tex]?
 
  • #6
They sure would. Don't forget dx*dy=r*dr*dtheta.
 
  • #7
Hmm, to check it, I plugged both expressions into my calculator and they're spitting out different numbers. The rectangular one gives me around 28.4 and the polar one gives me around 10.6.
 
  • #8
dtl42 said:
Hmm, to check it, I plugged both expressions into my calculator and they're spitting out different numbers. The rectangular one gives me around 28.4 and the polar one gives me around 10.6.

You've got a better calculator than I do. Did you catch "Don't forget dx*dy=r*dr*dtheta." I edited that into my last post.
 
  • #9
Ok, I've mastered my 'calculator'. I get 28.4 for both. Not exactly, of course.
 

What is a polar double integral?

A polar double integral is a type of integral that is used to calculate the volume or area of a region in polar coordinates. It involves integrating a function over a specific region in the polar plane.

How is a polar double integral different from a regular double integral?

A polar double integral is different from a regular double integral in that it uses polar coordinates instead of rectangular coordinates. This allows for simpler and more efficient calculations when dealing with circular or curved regions.

What is the formula for calculating a polar double integral?

The formula for calculating a polar double integral is ∫∫f(r,θ)rdrdθ, where f(r,θ) is the function being integrated, r is the radius, and θ is the angle.

What are some common applications of polar double integrals?

Polar double integrals are commonly used in physics and engineering to calculate the mass, center of mass, and moments of inertia of objects with circular or curved shapes. They are also used in calculating electric and magnetic fields in polar coordinates.

What are the limitations of polar double integrals?

Polar double integrals can only be used to calculate volume or area in regions that are symmetric about the origin. They also cannot handle regions with holes or disjointed regions. In these cases, a regular double integral or other methods may be needed.

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