Finding area of region enclosed by several curves in the first quadrant

Question

Find the area of the region in the first quadrant enclosed by the curves $$ xy = 4, xy = 10, y = e^4x, y = e^9x $$

Im aware i have to do a substitution, but Im not sure how to go about finding a good one.

Answer 1

Hint: you can do a change of variable adapted to the figure. With $$u = xy,\qquad v = y/x,$$ the integration limits in the new variables are constant and the jacobian is...

EDIT: about the jacobian: $$u/v = x^2\implies x =\sqrt{u/v},$$ $$uv = y^2\implies y =\sqrt{uv},$$ $$\frac{\partial(x,y)}{\partial(u,v)} = \cdots$$ Alternative possibility: $$ \frac{\partial(u,v)}{\partial(x,y)} = \left|\matrix{y&x\cr -y/x^2& 1/x}\right| = 2y/x = 2v $$ and $$\frac{\partial(x,y)}{\partial(u,v)} = \frac1{\frac{\partial(u,v)}{\partial(x,y)}} =\frac1{2v}.$$

Answer 2

Hint:

In polar coordinates we seek to find the area between$$r=\dfrac{\sqrt{10}}{\sqrt{\sin\theta\cos\theta}}\\r=\dfrac{2}{\sqrt{\sin\theta\cos\theta}}\\\theta=\tan^{-1}e^4\\\theta=\tan^{-1}e^9$$so we have$$S=\int_{\tan^{-1}e^4}^{\tan^{-1}e^9}\int_{\frac{\sqrt{10}}{\sqrt{\sin\theta\cos\theta}}}^{\frac{2}{\sqrt{\sin\theta\cos\theta}}}rdrd\theta=\int_{\tan^{-1}e^4}^{\tan^{-1}e^9}\dfrac{3}{\sin\theta\cos\theta}d\theta$$