Polar Double Integral: Converting and Evaluating

[dtl42]

Homework Statement

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

Homework Equations

r=\sqrt{(x^2+y^2)}

The Attempt at a Solution

Ok, I've gotten bounds on \theta, \frac{\pi}{6} \le \theta \le \frac{\pi}{3}. I'm not sure what the bounds for r should be, otherwise I have the integral:
\int_\frac{\pi}{6}^\frac{\pi}{3}\int_{?}^{?}{r^3}dr \; d\theta

 

[Dick]

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.

 

[dtl42]

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.

 

[Dick]

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.

 

[dtl42]

Wow, I'm sorry for that typo. That would change things. Would the correct limits for r be \frac{1}{cos(\theta)}\text{and}\frac{2}{cos(\theta)}?

 

[Dick]

They sure would. Don't forget dx*dy=r*dr*dtheta.

 

[dtl42]

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.

 

[Dick]

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.

 

[Dick]

Ok, I've mastered my 'calculator'. I get 28.4 for both. Not exactly, of course.