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Answer:
A.
Step-by-step explanation:
So to get the area of a square, we need to find the length of one side.
We know the length of the larger square is a, so the area of the larger cube is
We can find the length of a side of the smaller square by using pythagoreans theorem to find the hypotenuse of the triangle formed in the bottom left corner. The length of one side along the x axis is a - b, and the length of the other side, along the y-axis, is b.
We can plug it into pythagoreans theorem to get
(C represents the length of one side of the smaller square, and the hypotenuse of the triangle)
The area of the smaller triangle is C squared to the area of the smaller triangle is
To get the ratio of the smaller square in comparison to the larger square we divide the area of the smaller square by the area of the larger square.
So the ratio should be
By inspecting the integrand, the "obvious" choice for substitution would be
<em>u</em> = <em>y</em> + <em>x</em>
<em>v</em> = <em>y</em> - <em>x</em>
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Solving for <em>x</em> and <em>y</em>, we would have
<em>x</em> = (<em>u</em> - <em>v</em>)/2
<em>y</em> = (<em>u</em> + <em>v</em>)/2
in which case the Jacobian and its determinant are
The trapezoid <em>R</em> has two of its edges on the lines <em>x</em> + <em>y</em> = 8 and <em>x</em> + <em>y</em> = 9, so right away, we have 8 ≤ <em>u</em> ≤ 9.
Then for <em>v</em>, we observe that when <em>x</em> = 0 (the lowest edge of <em>R</em>), <em>v</em> = <em>y</em> ; similarly, when <em>y</em> = 0 (the leftmost edge of <em>R</em>), <em>v</em> = -<em>x</em>. So
-<em>x</em> ≤ <em>v</em> ≤ <em>y</em>
-(<em>u</em> - <em>v</em>)/2 ≤ <em>v</em> ≤ (<em>u</em> + <em>v</em>)/2
-<em>u</em> + <em>v</em> ≤ 2<em>v</em> ≤ <em>u</em> + <em>v</em>
-<em>u</em> ≤ <em>v</em> ≤ <em>u</em>
<em />
So, the integral becomes