Question

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.

Answer 1

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:

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

Performing other tedious calculations on computer, obtain the answer:

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