Question

In evaluating a double integral over a region D, a sum of iterated integrals was obtained as follows:

∬Df(x,y)dA=∫5_0∫(2/5)y_0 f(x,y)dxdy+∫7_5∫(7−y)_0 f(x,y)dxdy.

Sketch the region D and express the double integral as an iterated integral with reversed order of integration.

∫b_a∫g2(x)_g1(x) f(x,y)dydx

a= b=

g1(x)= g2(x)=

Answer 1

Answer

$a=0$, $b=2$

$g_1(x)=\frac{5x}{2}$,  $g_2(x)=7-x$

Step-by-step explanation:

Given that

$\int \int Df(x,y)dA=\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy+\int_5^7\int_0^{7-y} f(x,y)dxdy\; \cdots (i)$

For the term  $\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy$.

Limits for $x$ is from $x=0$ to $x=\frac {2y}{5}$ and for $y$ is from $y=0$ to $y=5$  and the region $D$, for this double integration is the shaded region as shown in graph 1.

Now, reverse the order of integration, first integrate with respect to $y$ then with respect to $x$ . So, the limits of $y$ become from $y=\frac{5x}{2}$ to $y=5$ and limits of $x$ become from $x=0$ to $x=2$ as shown in graph 2.

So, on reversing the order of integration, this double integration can be written as

$\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx\; \cdots (ii)$

Similarly, for the other term  $\int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy$.

Limits for $x$ is from $x=0$ to $x=7-y$ and limits for $y$ is from $y=5$ to $y=7$  and the region $D$, for this double integration is the shaded region as shown in graph 3.

Now, reverse the order of integration, first integrate with respect to $y$ then with respect to $x$ . So, the limits of $y$ become from $y=5$ to $y=7-x$ and limits of $x$ become from $x=0$ to $x=2$ as shown in graph 4.

So, on reversing the order of integration, this double integration can be written as

$\int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy=\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx\;\cdots (iii)$

Hence, from equations $(i)$, $(ii)$ and $(iii)$ , on reversing the order of integration, the required expression is

$\int \int Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx+\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx$

$\Rightarrow \int \int Df(x,y)dA=\int_0 ^2\left(\int _ {\frac {5x}{2}}^5 f(x,y)+\int _5 ^ {7-x} f(x,y)\right)dydx$

$\Rightarrow \int \int Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^{7-x} f(x,y)dydx\; \cdots (iv)$

Now, compare the RHS of the equation $(iv)$ with

$\int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)dydx$

We have,

$a=0, b=2, g_1(x)=\frac{5x}{2}$ and $g_2(x)=7-x$.

Answer 2

The required values are,

$a=0\\b=2\\g_1(x)=\frac{5x}{2} \\g_2(x)=7-x$

Double Integral:

Double integral is mainly used to find the surface area of a 2d figure, and it is denoted using ‘ ∫∫’. We can easily find the area of a rectangular region by double integration.

Given function is,

$\int \int_Df(x,y)dA=\int_{0}^{5}\int_{0}^{2y}f(x,y)dxdy+\int_{5}^{7}\int_{0}^{7-y}f(x,y)dxdy$

We can write the double integral as a single term by reversing the order as,

$\int_{a}^{b}\int_{g_1(x)}^{g_2(x)}f(x,y)dydx=\int_{0}^{2}\int_{\frac{5x}{2}}^{7-x}f(x,y)dydx)$

The region D is attached below.

Hence, we get the values,

$a=0\\b=2\\g_1(x)=\frac{5x}{2} \\g_2(x)=7-x$

Learn more about the topic Double Integral:

https://brainly.com/question/18568839