Change of Variables — Question 2

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Question 2

In the first quadrant, let RR be bounded by xy=1,xy=4,xy=1,xy=9.xy=1,\quad xy=4,\quad \frac{x}{y}=1,\quad \frac{x}{y}=9. Use u=xyu=xy and v=x/yv=x/y to evaluate ∬R(x/y)dA\iint_R(x/y)\,dA.

Tasks

  1. Show that the transformation is one-to-one on RR.

  2. Find the inverse Jacobian.

  3. Evaluate and check the result against the integrand range.

Original worksheet page 1: question and worked solution for 4-8-002
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Question 2 – Solution

Strategy. Product and ratio coordinates straighten the hyperbolas and rays into a rectangle.

Step 1: Region and inverse In the first quadrant, x=uv,y=uv,x=\sqrt{uv},\qquad y=\sqrt{\frac{u}{v}}, so positive (u,v)(u,v) determine a unique (x,y)(x,y). The image is 1≤u≤41\le u\le 4, 1≤v≤91\le v\le 9.

See the diagram in the original worksheet below.

Step 2: Jacobian Differentiating the inverse formulas gives |∂(x,y)∂(u,v)|=12v.\left|\frac{\partial(x,y)}{\partial(u,v)}\right| =\frac{1}{2v}. Therefore the factor x/y=vx/y=v cancels the vv in the denominator.

Step 3: Evaluate I=∫14∫19v(12v)dvdu=12(3)(8)=12.I=\int_1^4\int_1^9v\left(\frac 1{2v}\right)dv\,du =\frac 12(3)(8)=\boxed{12}.

Verification The region area is ∫14∫19(2v)−1dvdu=3ln⁡3\int_1^4\int_1^9(2v)^{-1}dv\,du=3\ln 3. Since 1≤x/y≤91\le x/y\le 9, the bounds 3ln⁡3≤I≤27ln⁡33\ln 3\le I\le 27\ln 3 hold; I=12I=12 lies between them.

Original worksheet page 2: question and worked solution for 4-8-002

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