Need help with double integrals and change of variables.

Question

I'm trying to calculate the integral of $(x^2 + y^2) \;dx\;dy$ where it is bounded by the positive region where the four curves intersect.

$x^2 - y^2 =2$

$x^2 - y^2 =8$

$x^2 + y^2 =8$

$x^2 + y^2 =16$

So I know I need to do a change of variables and polar looks like a good choice to me. This is where I get lost. So I use $x=r\cos t$, $y=r\sin t$ to change to polar, calculate the Jacobian from $x(r,t)$ and $y(r,t)$ Then do the integral $(x(r,t)^2+y(r,t)^2)$ (jacobian) $drdt$ ? And how do I figure out the limits of integration?

Answer 1

For a transformation of variables, the curve equations suggest something reasonably straightforward:

$$ u \ = \ x^2 + y^2 \ \ \rightarrow \ \ u = 8 \ , \ u = 16 \ \ ; \ \ v \ = \ x^2 - y^2 \ \ \rightarrow \ \ v = 2 \ , \ v = 8 \ \ ; $$

$$ A(2 \sqrt{2} \ , \ 0 ) \ \rightarrow \ (8 \ , \ 8) \ \ , \ \ B(2 \sqrt{3} \ , \ 2 ) \ \rightarrow \ (16 \ , \ 8) \ \ , $$ $$ C(3 \ , \ \sqrt{7} ) \ \rightarrow \ (16 \ , \ 2) \ \ , \ \ D(\sqrt{5} \ , \ \sqrt{3} ) \ \rightarrow \ (8 \ , \ 2) \ \ . $$

Our integration is now a rectangle in the $ \ uv-$ plane with the "circulation" from $ \ A \ $ around to $ \ D \ $ and back to $ \ A \ $ in the "clockwise", rather than the "counter-clockwise" direction.

The double integral is transformed as $$ \iint_{\mathbf{R}} \ (x^2 + y^2) \ dx \ dy \ \ \rightarrow \ \ \iint_{\mathbf{R^{\star}}} \ u \ · \ | \mathfrak{J} | \ \ du \ dv \ \ , $$

with $ \ \mathfrak{J} \ $ being the Jacobian matrix. It will be easier to compute the inverse Jacobian and find the determinant from that:

$$ \mathfrak{J}^{-1} \ = \ \left| \begin{array}{cc} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{array} \right| \ \ = \ \ \left| \begin{array}{cc} 2x & 2y \\ 2x & -2y \end{array} \right| \ \ = \ \ -8xy \ \ \Rightarrow \ \ \mathfrak{J} \ = \ -\frac{1}{8xy} \ \ , $$

the minus-sign indicating that the orientation of the region in the $ \ uv-$ plane is reversed from that in the $ \ xy-$ plane, as we remarked earlier. We can solve the set of variable transformation equations to obtain

$$ u \ = \ x^2 + y^2 \ \ , \ \ v \ = \ x^2 - y^2 \ \ \Rightarrow \ \ 2x^2 \ = \ u + v \ \ , \ \ 2y^2 \ = \ u - v $$ $$ \Rightarrow \ \ 4 x^2 y^2 \ = \ u^2 - v^2 \ \ \Rightarrow \ \ 2xy \ = \ \sqrt{u^2 - v^2} \ \ . $$

Our transformed integral is thus

$$ \int_2^8 \int_8^{16} \ u \ · \ \left| \ -\frac14 · \frac{1}{\sqrt{u^2 - v^2}} \ \right| \ \ du \ dv \ \ = \ \ \frac14 \ \int_2^8 \int_8^{16} \ \frac{u}{\sqrt{u^2 - v^2}} \ \ du \ dv $$

$$ = \ \ \frac14 \ \int_2^8 \ \sqrt{u^2 - v^2} \left \vert_8^{16} \right. \ \ dv \ \ = \ \ \frac14 \ \int_2^8 \left( \ \sqrt{16^2 - v^2} \ - \ \sqrt{8^2 - v^2} \ \right) \ \ dv $$

$$ = \ \ \frac14 · \frac12 \ \left[ \ v \sqrt{16^2 - v^2} \ + \ 16^2 · \arcsin \left( \frac{v}{16} \right) \ - \ v \sqrt{8^2 - v^2} \ - \ 8^2 · \arcsin \left( \frac{v}{8} \right) \ \right] \left \vert_2^8 \right. $$

$$ = \ \ 8 \sqrt{3} \ + \ \frac{16}{3} \pi \ - \ 0 \ - \ 4 \pi \ - \ \frac32 \sqrt{7} \ - \ 32 · \arcsin \left( \frac18 \right) \ + \ \frac12 \sqrt{15} \ + \ 8 · \arcsin \left( \frac14 \right) $$

$$ = \ \ 8 \sqrt{3} \ + \ \frac{4}{3} \pi \ - \ \frac32 \sqrt{7} \ - \ 32 \arcsin \left( \frac18 \right) \ + \ \frac12 \sqrt{15} \ + \ 8 \arcsin \left( \frac14 \right) $$

$$ \approx \ \ 13.8564 \ + \ 4.1888 \ - \ 3.9686 \ - \ 4.0105 \ + \ 1.9365 \ + \ 2.0214 \ \ \approx \ \ \mathbf{14.0240} \ \ . $$

The two integral calculations in Cartesian and polar coordinates confirm this result.

$$ \ \ $$

$ \mathbf{APPENDIX -} $ We would like to verify that all three exact expressions are in fact equivalent. From the three calculations, we obtained

$ \mathbf{Cartesian :} \quad -\frac32 \sqrt{7} \ + \ \frac12 \sqrt{15} \ + \ 8 \sqrt{3} \ + \ \frac{40}{3} \pi \ - \ 64 \arcsin \left( \frac34 \right) \ + \ 16 \arcsin \left( \frac{\sqrt{5}}{2 \sqrt{2}} \right) $

$ \mathbf{polar :} \quad 8 \sqrt{3} \ - \ \frac{32}{3} \pi \ - \ 16 \arctan \left( \sqrt{\frac35} \right) + \ 64 \arctan \left( \frac{\sqrt{7}}{3} \right) + \ \frac12 \sqrt{15} \ - \frac32 \sqrt{7} $

$ \mathbf{uv-plane :} \quad 8 \sqrt{3} \ + \ \frac{4}{3} \pi \ - \ \frac32 \sqrt{7} \ - \ 32 \arcsin \left( \frac18 \right) \ + \ \frac12 \sqrt{15} \ + \ 8 \arcsin \left( \frac14 \right) $

The three "square-root" terms appear in all of these, so we only need to examine the trigonometric terms.

In comparing the Cartesian and polar terms, we note that $$ \alpha \ = \ \arcsin \left( \frac34 \right) \ = \ \arctan \left( \frac{3}{\sqrt{7}} \right) \ \ \text{and} \ \ \beta \ = \ \arcsin \left( \sqrt{\frac58} \right) \ = \ \arctan \left( \sqrt{\frac53} \right) \ \ . $$ So $ \ \frac{\pi}{2} - \alpha \ = \ \arctan \left( \frac{\sqrt{7}}{3} \right) \ $ and $ \ \frac{\pi}{2} - \beta \ = \ \arctan \left( \sqrt{\frac35} \right) \ \ . $ The three terms of interest in the polar-integral expression are then

$$ - \ \frac{32}{3} \pi \ - \ 16 · \left[ \frac{\pi}{2} - \beta \right] + \ 64 · \left[ \frac{\pi}{2} - \alpha \right] \ = \ \left( \frac{-32-24+96}{3} \right) \pi \ + \ 16 · \beta \ - \ 64 · \alpha $$

$$ = \ \frac{40}{3} \pi \ - \ 64 · \alpha \ + \ 16 · \beta \ \ ; $$

hence, the Cartesian and polar expressions are equivalent.

Checking the $ \ uv-$ plane result is somewhat tricky, in that it isn't immediately obvious what terms in the expressions should be "matched up". It turns out that for the Cartesian and $ \ uv-$ plane expressions that a "double-angle" with the same "added angle" works out:

$$ \frac{40}{3} \pi \ - \ 64 · \alpha \ + \ 16 · \beta \ \ = \ \ \frac{4}{3} \pi \ - \ 32 · ( 2 \alpha + \phi) \ + \ 8 · ( 2 \beta + \phi) $$ $$ \Rightarrow \ \ \frac{36}{3} \pi \ \ = \ -32 \phi \ + \ 8 \phi \ \ \Rightarrow \ \ 12 \pi \ = \ -24 \phi \ \ \Rightarrow \ \ \phi \ = \ -\frac{\pi}{2} \ \ . $$

Since we wish to compare sine values, we will need $$ \ \sin \left(2 \theta - \frac{\pi}{2} \right) \ = \ \sin (2 \theta) · \cos \left( \frac{\pi}{2} \right) \ - \ \cos (2 \theta) · \sin \left( \frac{\pi}{2} \right) \ = \ - \cos (2 \theta) \ = \ \sin^2 \theta \ - \ \cos^2 \theta \ \ . $$

From the statements above, we have $$ \sin \alpha \ = \ \frac34 \ , \ \cos \alpha \ = \ \frac{\sqrt{7}}{4} \ , \ \sin \beta \ = \ \sqrt{\frac58} \ , \ \sin \beta \ = \ \sqrt{\frac38} \ \ , $$

hence,

$$ \ \sin \left(2 \alpha - \frac{\pi}{2} \right) \ = \ \left( \frac34 \right)^2 \ - \ \left(\frac{\sqrt{7}}{4} \right)^2 \ = \ \frac{9 - 7}{16} \ \ = \ \ \frac18 \ \ \text{and} $$ $$ \ \sin \left(2 \beta - \frac{\pi}{2} \right) \ = \ \left( \sqrt{\frac58} \right)^2 \ - \ \left(\sqrt{\frac38} \right)^2 \ = \ \frac{5 - 3}{8} \ \ = \ \ \frac14 \ \ . $$

With the correspondences indicated between the angles in each expression, we conclude that the exact expressions found for the double integral by each method are indeed equivalent.

Answer 2

I had a very similar problem when I took calculus. Try this:

Let $ u = x^2 - y^2 $ $ v = x^2 + y^2 $

Now your region of integration is v from 8 to 16 and u from 2 to 8. Now evaluate your Jacobian and throw it into the integral. The integrand in your problem...check it out... is now just u! Sweet. That's what we want. Your Jacobian is the only tricky part, but you'll get it.

When dealing with problems like this, a really good question to ask yourself is, "how can I make my limits of integration easier?" For instance, sometimes converting to polar makes the limits really easy, but keep your mind open to new substitutions, especially when you don't have to integrate from zero to r or some other variable or function. The best possible case is when you can get both of your regions in terms of real values, as we did by letting $ u = x^2 - y^2 $ $ v = x^2 + y^2 $ Now the only tricky part is the Jacobian. If you want a hint with that, let me know.

Answer 3

Since this problem was never fully answered, I decided to use it as an illustration of different integration approaches that can be taken with the double integral. The region of interest is one of four bounded by two circles and two "horizontal" hyperbolas, all concentric on the origin, due to the "four-fold" symmetry of the curves. We are to work with the region found in the first quadrant; the graph above identifies the bounding curves and the "vertices" of the region.

These vertices are found from pairing equations of each circle with those of each hyperbola to produce, for example,

$$ x^2 - y^2 = 2 \ , \ x^2 + y^2 = 8 \ \ \Rightarrow \ \ x^2 = 5 \ , \ y^2 = 3 \ \ \Rightarrow \ \ D (\sqrt{5} \ , \ \sqrt{3}) \ \ . $$

The other three are determined similarly: $ \ A(2 \sqrt{2} \ , \ 0) \ , \ C(3 \ , \ \sqrt{7}) \ , \ B(2 \sqrt{3} \ , \ 2) \ \ . $

In this answer, the integral will be calculated in Cartesian coordinates, which requires the greatest amount of effort. We will require three separate integrals between the various bounding curves:

$$ \iint_{\mathbf{R}} \ (x^2 + y^2) \ \ dx \ dy \ \ = \ \ \int_{\sqrt{8 - x^2}}^{\sqrt{x^2 - 2}} \int_{\sqrt{3}}^{2 \sqrt{2}} \ (x^2 + y^2) \ \ dx \ dy $$ $$ + \ \ \int_{\sqrt{x^2 - 8}}^{\sqrt{x^2 - 2}} \int^{\sqrt{7}}_{2 \sqrt{2}} \ (x^2 + y^2) \ \ dx \ dy \ \ + \ \ \int_{\sqrt{x^2 - 8}}^{\sqrt{16 - x^2}} \int_{\sqrt{7}}^{2 \sqrt{3}} \ (x^2 + y^2) \ \ dx \ dy $$

$$ = \ \ \int_{\sqrt{3}}^{2 \sqrt{2}} \left(x^2y \ + \ \frac13 y^3 \right) \left\vert_{\sqrt{8 - x^2}}^{\sqrt{x^2 - 2}} \right. \ \ dx \ \ + \ \ \int^{\sqrt{7}}_{2 \sqrt{2}} \left(x^2y \ + \ \frac13 y^3 \right) \left\vert_{\sqrt{x^2 - 8}}^{\sqrt{x^2 - 2}} \right.\ \ dx $$ $$ + \ \ \int_{\sqrt{x^2 - 8}}^{\sqrt{16 - x^2}} \int_{\sqrt{7}}^{2 \sqrt{3}} \left(x^2y \ + \ \frac13 y^3 \right) \left\vert_{\sqrt{x^2 - 8}}^{\sqrt{16 - x^2}} \right. \ \ dx $$

$$ = \ \ \int_{\sqrt{3}}^{2 \sqrt{2}} \left(x^2\sqrt{x^2 - 2} \ + \ \frac13 (x^2 - 2)^{3/2} \ - \ x^2\sqrt{8 - x^2} \ - \ \frac13 (8 - x^2)^{3/2} \right) \ \ dx $$ $$ + \ \ \int^{\sqrt{7}}_{2 \sqrt{2}} \left(x^2\sqrt{x^2 - 2} \ + \ \frac13 (x^2 - 2)^{3/2} \ - \ x^2\sqrt{x^2 - 8} \ - \ \frac13 (x^2 - 8)^{3/2} \right) \ \ dx $$ $$ + \ \ \int_{\sqrt{7}}^{2 \sqrt{3}} \left(x^2\sqrt{16 - x^2} \ + \ \frac13 (16 - x^2)^{3/2} \ - \ x^2\sqrt{x^2 - 8} \ - \ \frac13 (x^2 - 8)^{3/2} \right) \ \ dx \ \ . $$

With the aid of the indefinite integrals

$$ \int \ x^2 \sqrt{x^2 - a^2} \ \ dx \ \ = \ \ \frac18 x \ (2x^2 - a^2) \ \sqrt{x^2 - a^2} \ - \ \frac18 \ a^4 \ \log \left(\sqrt{x^2 - a^2} \ + \ x \right) \ + \ C \ \ , $$ $$ \int \ x^2 \sqrt{a^2 - x^2} \ \ dx \ \ = \ \ \frac18 x \ (2x^2 - a^2) \ \sqrt{a^2 - x^2} \ + \ \frac18 \ a^4 \ \arcsin \left( \frac{x}{a} \right) \ + \ C \ \ , $$ $$ \int \ (x^2 - a^2)^{3/2} \ \ dx \ \ = \ \ \frac18 x \ (2x^2 - 5a^2) \ \sqrt{x^2 - a^2} \ + \ \frac38 \ a^4 \ \log \left(\sqrt{x^2 - a^2} \ + \ x \right) \ + \ C \ \ , $$ $$ \int \ (a^2 - x^2)^{3/2} \ \ dx \ \ = \ \ -\frac18 x \ (2x^2 - 5a^2) \ \sqrt{a^2 - x^2} \ + \ \frac38 \ a^4 \ \arcsin \left( \frac{x}{a} \right) \ + \ C \ \ , $$

and a certain amount of cancelation and simplification, the sum of the three definite integrals is evaluated as

$$ \iint_{\mathbf{R}} \ (x^2 + y^2) \ \ dx \ dy \ \ = \ \ -\frac32 \sqrt{7} \ + \ \frac12 \sqrt{15} \ + \ 8 \sqrt{3} \ + \ \frac{40}{3} \pi \ - \ 64 \arcsin \left( \frac34 \right) $$ $$ + \ 16 \arcsin \left( \frac{\sqrt{5}}{2 \sqrt{2}} \right) \ \ \approx \ -3.9686 \ + \ 1.9365 \ + \ 13.8564 \ + \ 41.8879 \ - \ 54.2760 \ + \ 14.5878 \ \ \approx \ \ \mathbf{14.0240} \ \ . $$

The other answers I have added evaluate the integral over this region using polar coordinates and by transforming the region into a rectangle.

Answer 4

Because the boundary circles and hyperbolas for the integration region are concentric on the origin, the use of polar coordinates simplifies the necessary work considerably. Here we have

$$ x^2 + y^2 \ = \ a^2 \ \ \rightarrow \ \ r \ = \ a \ \ , $$ $$ x^2 - y^2 \ = \ a^2 \ \ \rightarrow \ \ r^2 (\cos^2 \theta - \sin^2 \theta) \ = \ a^2 \ \ \Rightarrow \ \ r^2 \ = \ \frac{a^2}{\cos^2 (2 \theta)} \ = \ a^2 \sec^2 (2 \theta) \ \ , $$

$$ A(2 \sqrt{2} \ , \ 0) \ \rightarrow \ (2 \sqrt{2} \ , \ 0) \ \ , \ \ B(2 \sqrt{3} \ , \ 2) \ \rightarrow \ \left(4 \ , \ \frac{\pi}{6} \right) \ \ , $$ $$ D( \sqrt{5} \ , \ \sqrt{3}) \ \rightarrow \ (2 \sqrt{2} \ , \ \Theta) \ \ , \ \ C(3 \ , \ \sqrt{7}) \ \rightarrow \ (4 \ , \ \Phi ) \ \ , $$

where $ \ \tan \Theta \ = \ \sqrt{\frac35} \ , \ \cos \Theta \ = \ \sqrt{\frac58} \ , \ \sin \Theta \ = \ \sqrt{\frac38} \ $ and $ \ \tan \Phi \ = \ \frac{\sqrt{7}}{3} \ , \ \cos \Phi \ = \ \frac34 \ , \ \sin \Phi \ = \ \frac{\sqrt{7}}{4} \ \ . $

The integration in polar coordinates must also cover three regions, so we have $$ \iint_{\mathbf{R}} \ (x^2 + y^2) \ dx \ dy \ \ \rightarrow \ \iint_{\mathbf{R'}} \ r^2 \ \ r \ dr \ d\theta \ : $$

$$ \int_0^{\pi/6} \int_{2 \sqrt{2}}^{\sqrt{8 \sec(2\theta)}} \ r^3 \ dr \ d\theta \ + \ \int^{\Theta}_{\pi/6} \int_{2 \sqrt{2}}^{4} \ r^3 \ dr \ d\theta \ + \ \int_{\Theta}^{\Phi} \int^{4}_{\sqrt{2 \sec(2\theta)}} \ r^3 \ dr \ d\theta $$

$$ = \ \ \frac14 \ \int_0^{\pi/6} \ r^4 \left \vert_{2 \sqrt{2}}^{\sqrt{8 \sec(2\theta)}} \right. \ d\theta \ + \ \frac14 \ \int^{\Theta}_{\pi/6} \ r^4 \left \vert_{2 \sqrt{2}}^{4} \right. \ d\theta \ + \ \frac14 \ \int_{\Theta}^{\Phi} \ r^4 \left \vert^{4}_{\sqrt{2 \sec(2\theta)}} \right. \ d\theta $$

$$ = \ \ \frac14 \ \int_0^{\pi/6} \ [64 \sec^2 (2\theta) \ - \ 64] \ \ d\theta \ + \ \frac14 \ \int^{\Theta}_{\pi/6} \ [256 \ - \ 64] \ \ d\theta $$ $$ + \ \frac14 \ \int_{\Theta}^{\Phi} \ [256 \ - \ 4 \sec^2 (2\theta) ] \ \ d\theta $$

$$ = \ \ \frac18 · \ [64 \tan (2\theta) \ - \ 64 · \theta] \left\vert_0^{\pi/6} \right. \ \ d\theta \ + \ \frac18 · 192 · \theta \left\vert^{\Theta}_{\pi/6} \right. \ d\theta \ + \ \frac18 · \ [256 · \theta \ - \ 4 \tan (2\theta) ] \left\vert_{\Theta}^{\Phi} \right. \ \ d\theta $$

$$ = \ \ldots \ = \ \ 8 \sqrt{3} \ - \ \frac{32}{3} \pi \ - \ 16 \arctan \left( \sqrt{\frac35} \right) + \ 64 \arctan \left( \frac{\sqrt{7}}{3} \right) + \left( \frac{\tan \Theta}{1 - \tan^2 \Theta} \right) - \left( \frac{\tan \Phi}{1 - \tan^2 \Phi} \right) $$

$$ = \ \ 8 \sqrt{3} \ - \ \frac{32}{3} \pi \ - \ 16 \arctan \left( \sqrt{\frac35} \right) + \ 64 \arctan \left( \frac{\sqrt{7}}{3} \right) + \ \frac12 \sqrt{15} \ - \frac32 \sqrt{7} $$

$$ \approx \ \ 13.8564 \ - \ 33.5103 \ - \ 10.5449 \ + \ 46.2550 \ + \ 1.9365 \ - \ 3.9686 \ \ \approx \ \ \mathbf{14.0241} \ \ . $$

In the third answer, we'll use a variable transformation to a rectangular region; the other methods shown are intended to confirm the value of the double integral that we find there.