Answer to Question #97153 in Calculus for Rachel

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

Solution:

We are going to evaluate the integral bounded by the region D.

D is the given region bounded by

$$XY =a, XY = b, XY^{1.4} = c \space and \space XY^ {1.4} = d$$

where 0>a>b and 0<c<d.

Choose

$$XY = U \space and \space XY^{1.4} = V$$

( Here, $U = a, U = b \space and \space V= c \space and \space V = d)$

Now, divide V by U

$$\frac {V}{U} =\frac {XY^{1.4}} {XY} = Y^{0.4}$$

We can write this equation for Y,

$$Y = (\frac {V}{U})^{\frac {10} {4}} = (\frac {V}{U})^{2.5}$$

$$Y^{10} = (\frac{V}{U})^{25}$$

Now we can solve for X,

$$X = \frac {U}{V} = \frac {U}{(\frac {V} {U})^{2.5}} = \frac {U \times U^{2.5}} {V^{2.5}} = \frac {U^{3.5}}{V^{2.5}}$$

$$X^8 = (\frac {U^{3.5}}{V^{2.5}})^ 8 = \frac {U^{28}}{V^{20}}$$

Now,

$$X^8 \times Y^{10} = \frac { U^{28}} {V^{20}} \times \frac {V^{25}}{U^{25}}= U^3 V^5$$

$$dA = J \space dv \times du$$

Here,

$$J = \frac {\partial (X,Y)}{\partial (U,V)} =\begin{vmatrix} \frac {\partial X}{\partial U} & \frac {\partial X}{\partial V} \\ \frac {\partial Y}{\partial U} & \frac {\partial Y}{\partial V} \end{vmatrix} = \frac {8.75}{V} - \frac {6.25}{V} = \frac {2.5} {V}$$

$$\iint _D X^8 Y^{10} dA = \int_{U=a}^{U =b} \int _{V=c} ^{V=d} U^3 V^5 \frac {5}{2V} dV dU$$

=

$$= \frac {5}{2} \int_{U=a}^{U =b} \int _{V=c} ^{V=d} U^3 V^4 dV dU = \frac {5}{2} \int _{U=a} ^{U=b} U^3 [\frac {V^5}{5}]_c^ d dU$$

$$=\frac {1}{2} (d^5 - c^5 ) [\frac {U^4}{4}]_a^b = \frac {1 }{8} (b^4 - a^4) (d^5 - c^5)$$

Answer:

Required Integral =

$$\frac {1 }{8} (b^4 - a^4) (d^5 - c^5)$$