Answer to Question #96327 in Calculus for Rachel

Question #96327
Evaluate the following integral if D is the region bounded by xy=a, xy=b, xy^1.4=c and xy^1.4=d, where 0< a < b and 0< c < d. ∫ ∫ D x^8y^10 dA. Note: D should be at the bottom of the second integral, it is not part of the function.
1
Expert's answer
2019-10-15T10:14:59-0400

Let's visualize the area of integration (filled region):



The left (yellow) curve stands for "xy=a", the right (red) one for "xy=b", the top (green) one for "xy^{1.4}=d" and the bottom (blue) curve is for "xy^{1.4}=c".

Now let's define the limits of integration.

The most left point is intercection between "xy=a" and "xy^{1.4}=d" curves. So "y_1 = \\dfrac{a}{x} = (\\dfrac{d}{x})^{5\/7}\\\\" . Expressing x:

"x_1 = \\dfrac{a^{7\/2}}{d^{5\/2}}" . Using the similar reasoning, find ordinates of vertices in order from left to right:

"x_2 = \\dfrac{a^{7\/2}}{c^{5\/2}}" , "x_3 = \\dfrac{b^{7\/2}}{d^{5\/2}}" , "x_4 = \\dfrac{b^{7\/2}}{c^{5\/2}}\\\\".

Now let's divide our region in three subregions and calculate integlal over each of them.



The integral over first subregion:

"I_I = \\int_{x_1}^{x_2} \\limits dx \\int_{\\frac{a}{x}}^{(\\frac{d}{x})^{5\/7}} \\limits x^8y^{10} dy = \n\\dfrac{1}{11} \\int_{x_1}^{x_2} \\limits dx \\cdot x^8[ ( \\frac{d}{x} )^{55\/7} - \\frac{a}{x} )^{11}] = \n \\dfrac{1}{11} \\int_{x_1}^{x_2} \\limits dx\\cdot x^{1\/7} d^{55\/7} - \n\\dfrac{1}{11} \\int_{x_1}^{x_2} \\limits dx\\cdot a^{11}x^{-3} = \n\\dfrac{d^{55\/7}}{11} \\cdot \\dfrac{7}{8}\\cdot x^{8\/7}\\Bigg|_{x1}^{x2} + \n\\dfrac{a^{11}}{11} \\cdot \\dfrac{1}{2x^2}\\Bigg|_{x1}^{x2} = \n-\\dfrac{7d^{55\/7}a^4}{88d^{20\/7}} + \\dfrac{7d^{55\/7}a^4}{88c^{20\/7}} -\n\\dfrac{a^{11}d^5}{22a^7} + \\dfrac{a^{11}c^5}{22a^7} = \n-\\dfrac{7d^{35\/7}a^4}{88} + \\dfrac{7d^{35\/7}a^4}{88c^{20\/7}} - \n\\dfrac{a^{4}d^5}{22} + \\dfrac{a^{4}c^5}{22} \n\n\\\\"

Performing other tedious calculations on computer, obtain the answer:

"A =\\dfrac{1}{44 a^7}\\cdot \\left( a^{21\/2} \\left(\\frac{7 d^{55\/7} \\sqrt[7]{\\frac{a^{7\/2}}{c^{5\/2}}}}{c^{5\/2}}-7\n d^{75\/14} \\sqrt[7]{\\frac{a^{7\/2}}{d^{5\/2}}}\\right)+2 a^{11} \\left(c^5-d^5\\right)+2 b^{11}\n \\left(c^5-d^5\\right) \\right)"


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