The double integral $\underset{0}{\overset{a}{\int }}\underset{0}{\overset{y}{\int }}f\left(x,y\right)$ dx dy is equivalent to

Understanding the Original Double Integral

The problem asks us to find an equivalent form of the double integral:

`$\underset{0}{\overset{a}{\int }}\underset{0}{\overset{y}{\int }}f\left(x,y\right)dxdy$ `

This is an iterated integral, meaning we integrate with respect to one variable at a time. Let's break down the limits:

- The inner integral is `$\underset{0}{\overset{y}{\int }}f\left(x,y\right)dx$ . This means for a fixed value of $y$ , we integrate $f(x,y)$ with respect to $x$ , where $x$ ranges from 0 to $y$ . So, we have the condition $0\le x\le y$ `.
- The outer integral is `$\underset{0}{\overset{a}{\int }}...dy$ . This means we integrate the result of the inner integral with respect to $y$ , where $y$ ranges from 0 to $a$ . So, we have the condition $0\le y\le a$ `.

Combining these conditions, the region of integration `$R$ ` is defined by:

`$R=\{(x,y)\mid 0\le x\le y,0\le y\le a\}$ `

Visualizing this region in the xy-plane:

- `$y=x$ ` is a line passing through the origin with a slope of 1.
- `$x=0$ ` is the y-axis.
- `$y=a$ ` is a horizontal line.
- `$y=0$ ` is the x-axis.

The region is a triangle bounded by the y-axis (`$x=0$ ), the line $y=a$ , and the line $y=x$ `. The vertices of this triangle are (0, 0), (0, a), and (a, a).

Changing the Order of Integration

We want to change the order of integration from `$dxdy$ to $dydx$ . This means the outer integral will be with respect to $x$ and the inner integral with respect to $y$ `.

To do this, we need to describe the same region `$R$ using bounds for $y$ first, and then bounds for $x$ `.

- First, determine the range of `$x$ values in the region. Looking at the vertices (0, 0), (0, a), and (a, a), the smallest $x$ value is 0 and the largest is $a$ . So, the outer limits for $x$ will be from 0 to $a$ , i.e., $0\le x\le a$ `.
- Next, for a fixed value of `$x$ within this range, determine the range of $y$ values. Imagine a vertical line segment at a specific $x$ within the region. This line enters the region at the line $y=x$ and exits the region at the line $y=a$ . So, for a fixed $x$ , $y$ ranges from $x$ to $a$ , i.e., $x\le y\le a$ `.

With these new bounds, the equivalent double integral with the order changed to `$dydx$ ` is:

`$\underset{0}{\overset{a}{\int }}\left(\underset{x}{\overset{a}{\int }}f(x,y)dy\right)dx$ `

Note the change in the inner differential order from `$dx$ to $dy$ and the outer differential order from $dy$ to $dx$ `.

Matching the Equivalent Integral

Now, let's compare our derived integral with the options provided:

- Option 1: `$\underset{0}{\overset{x}{\int }}\underset{0}{\overset{y}{\int }}f\left(x,y\right)dxdy$ ` - Incorrect limits and order.
- Option 2: `$\underset{0}{\overset{a}{\int }}\underset{x}{\overset{y}{\int }}f\left(x,y\right)dxdy$ ` - Incorrect order and inner limit.
- Option 3: `$\underset{0}{\overset{a}{\int }}\underset{x}{\overset{a}{\int }}f\left(x,y\right)dydx$ ` - This matches our derived integral exactly.
- Option 4: `$\underset{0}{\overset{a}{\int }}\underset{0}{\overset{a}{\int }}f\left(x,y\right)dxdy$ ` - Incorrect limits, defines a square region.

Therefore, the equivalent double integral is given by Option 3.