Find the real number a for which the value of theiterated integralV1- (y-.(– 2² – ỷ + 2y) dx dy- V1- (y-a)is a maximum, and find the maximum value.a-1

Answered: Find the real number a for which the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Find the real number a for which the value of the
iterated integral
V1- (y-.
(– 2² – ỷ + 2y) dx dy
- V1- (y-a)
is a maximum, and find the maximum value.
a-1

Expert Solution

Step 1

Let $f (x, y) = - x^{2} - y^{2} + 2 y$ . To determine the maximum value of the function, differentiate f(x,y) partially with respect to x and y and then solve the equations $f_{x} = 0$ and $f_{y} = 0$ for x and y.

$\begin{array}{rcl} f (x, y) & = & - x^{2} - y^{2} + 2 y \\ f_{x} & = & - 2 x \\ f_{y} & = & - 2 y + 2 \\ f_{x x} & = & - 2 \\ f_{y y} & = & - 2 \\ f_{x y} & = & 0 \\ f_{x} & = & 0 \\ - 2 x & = & 0 \\ x & = & 0 \\ f_{y} & = & 0 \\ - 2 y + 2 & = & 0 \\ - 2 y & = & - 2 \\ y & = & 1 \\ D (x, y) & = & f_{x x} f_{y y} - f_{x y}^{2} \\ D (0, 1) & = & (- 2) (- 2) - (0) \\ = & 4 \\ > & 0 \\ f_{x x} (0, 1) & = & - 2 \\ < & 0 \end{array}$

Therefore, the function f(x,y) is maximum, when $x = 0$ and $y = 1$ .

Step 2

To obtain the value of a, substitute 0 for x and 1 for y in the limits of integration, and simplify

$\begin{array}{rcl} x & = & \sqrt{1 - {(y - a)}^{2}} \\ 0 & = & \sqrt{1 - {(1 - a)}^{2}} \\ = & 1 - {(1 - a)}^{2} \\ {(1 - a)}^{2} & = & 1 \\ 1 - a & = & 1 \\ 1 - 1 & = & a \\ 0 & = & a \\ y & = & a - 1 \\ 1 & = & a - 1 \\ a & = & 2 \end{array}$

Hence, $a = 0$ and $a = 2$ .

Step 3

To find the maximum value of the integral, substitute 0 for a in the integral $\int_{y = a - 1}^{a + 1} \int_{x = - \sqrt{1 - {(y - a)}^{2}}}^{\sqrt{1 - {(y - a)}^{2}}} (- x^{2} - y^{2} + 2 y) d x d y$ , and evaluate it.

$\begin{array}{rcl} \int_{y = a - 1}^{a + 1} \int_{x = - \sqrt{1 - {(y - a)}^{2}}}^{\sqrt{1 - {(y - a)}^{2}}} (- x^{2} - y^{2} + 2 y) d x d y & = & \int_{y = 0 - 1}^{0 + 1} \int_{x = - \sqrt{1 - {(y - 0)}^{2}}}^{\sqrt{1 - {(y - 0)}^{2}}} (- x^{2} - y^{2} + 2 y) d x d y \\ = & \int_{y = - 1}^{1} \int_{x = - \sqrt{1 - y^{2}}}^{\sqrt{1 - y^{2}}} (- x^{2} - y^{2} + 2 y) d x d y \\ = & \int_{y = - 1}^{1} {[- (\frac{x^{3}}{3}) - x y^{2} + 2 x y]}_{x = - \sqrt{1 - y^{2}}}^{\sqrt{1 - y^{2}}} d y \\ = & \int_{y = - 1}^{1} ([- \frac{1}{3} {(1 - y^{2})}^{\frac{3}{2}} - {(1 - y^{2})}^{\frac{1}{2}} y^{2} + 2 {(1 - y^{2})}^{\frac{1}{2}} y] - [\frac{1}{3} {(1 - y^{2})}^{\frac{3}{2}} + {(1 - y^{2})}^{\frac{1}{2}} y^{2} - 2 {(1 - y^{2})}^{\frac{1}{2}} y]) d y \\ = & \int_{y = - 1}^{1} [- \frac{2}{3} {(1 - y^{2})}^{\frac{3}{2}} - 2 {(1 - y^{2})}^{\frac{1}{2}} y^{2} + 4 {(1 - y^{2})}^{\frac{1}{2}} y] d y \\ = & - \frac{2}{3} \int_{y = - 1}^{1} {(1 - y^{2})}^{\frac{3}{2}} d y - 2 \int_{y = - 1}^{1} y^{2} {(1 - y^{2})}^{\frac{1}{2}} d y + 4 \int_{y = - 1}^{1} y {(1 - y^{2})}^{\frac{1}{2}} d y \\ = & - \frac{2}{3} I_{1} - 2 I_{2} + 4 I_{3} \dots (1) \end{array}$

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