Question

Write an iterated integral of a continuous function \(f\) over the following regions. The region in the first quadrant bounded by the \(x\) -axis, the line \(x=6-y,\) and the curve \(y=\sqrt{x}\)

Short answer

Question: Find the iterated integral of a continuous function \(f(x, y)\) over a region bounded by the x-axis, the line \(x=6-y\), and the curve \(y=\sqrt{x}\). Solution: The iterated integral of the function \(f(x, y)\) over the given region is: $$ \int_{0}^{2} \int_{6-y}^{\sqrt{y^2}} f(x, y) \, dx \, dy $$

Step-by-step solution

Find the intersection points of the curves

To find the intersection points of the given curves, we need to solve the following systems of equations: 1. Intersection of \(x=6-y\) and \(y=\sqrt{x}\): Solve for \(y\), and then find the corresponding \(x\)-values. 2. Intersection of \(x=6-y\) and the \(x\)-axis: Since the \(x\)-axis is represented by the equation \(y=0\), we only need to solve for \(x\) when \(y=0\) in the equation \(x=6-y\). 3. Intersection of \(y=\sqrt{x}\) and the \(x\)-axis: Same as, find the point where the curve \(y=\sqrt{x}\) meets the \(x\)-axis.

Calculate the intersection points

1. Intersection between \(x=6-y\) and \(y=\sqrt{x}\): To find the intersection, we substitute \(y\) from one equation into the other: $$ y = \sqrt{6-y} $$ Square both sides: $$ y^2 = 6-y $$ Rearrange to get a quadratic equation: $$ y^2 + y -6 = 0 $$ Factor the quadratic: $$ (y+3)(y-2) = 0$$ So, we have two solutions for y: \(y=-3\) and \(y=2\). Since we are only considering the first quadrant, we will consider only \(y=2\). The corresponding x-value is: $$ x = 6 - y = 6 - 2 = 4 $$ So, the intersection point between the two curves is \((4,2)\). 2. Intersection between the \(x=6-y\) line and \(x\)-axis: When \(y = 0\), \(x = 6 - 0 = 6\). Therefore, the intersection with the x-axis is \((6,0)\). 3. Intersection between the \(y=\sqrt{x}\) curve and \(x\)-axis: When \(y = 0\), \(x = 0\). Therefore, the intersection with the x-axis is \((0,0)\).

Write the iterated integral with the appropriate limits of integration

Now that we have determined the points of intersection, we are ready to write the iterated integral. We will perform the \(x\)-integration first and then the \(y\)-integration. The iterated integral for the given function \(f(x, y)\) is as follows: $$ \int_{0}^{2} \int_{6-y}^{\sqrt{y^2}} f(x, y) \, dx \, dy $$

Key concepts

Intersection Points

Intersection points are crucial when setting up the limits for an iterated integral. They represent specific locations where two curves or lines meet on a coordinate plane. In our example, we need to find where the line \(x = 6 - y\) intersects both the curve \(y = \sqrt{x}\) and the \(x\)-axis.
1. **Intersection of \(x = 6 - y\) and \(y = \sqrt{x}\):** We substitute one equation into the other to solve for \(y\). By setting \(y = \sqrt{6-y}\), squaring both sides, and rearranging terms, we develop the quadratic \(y^2 + y - 6 = 0\). Solving this, the feasible solution (since we are in the first quadrant) is \(y = 2\) which results in \(x = 4\). The intersection point is \((4, 2)\).
2. **Intersection of \(x = 6 - y\) with the \(x\)-axis:** Here, \(y\) must be zero, leading to \(x = 6\). Thus, the intersection with the \(x\)-axis is at \((6, 0)\).
3. **Intersection of \(y = \sqrt{x}\) with the \(x\)-axis:** Setting \(y = 0\) gives \(x = 0\). This intersection is simply \((0, 0)\).
Understanding these intersections gives us the information needed to set proper bounds for integration.

Limits of Integration

Once you have identified the intersection points, it's time to define the limits of integration for the iterated integral, which dictate the region over which the integration occurs.

• **Horizontal or \(x\)-limits:** These limits define the right and left bounds of the region at any specific \(y\). Observing our problem, the right limit is \(x = \sqrt{y^2}\), reflecting the curve \(y = \sqrt{x}\) when reversed to solve for \(x\). The left limit, starting at \(6 - y\), comes from expressing \(x\) in terms of \(y\) from the line equation.

• **Vertical or \(y\)-limits:** These run vertically from the lowest to the highest intersections in the region. Here, \(y\) spans from \(0\) to \(2\), discovered from the solutions when \(y = \sqrt{x}\) and \(x = 6 - y\) were solved.

Overall, the integration happens from the intersection at \((0,0)\) or \((6,0)\) to the point \((4,2)\). The integral notation integrates first over \(x\), then over \(y\), resulting in an expression like \[\int_{0}^{2} \int_{6-y}^{\sqrt{y^2}} f(x, y) \, dx \, dy\]. This systematic approach ensures the entire region in the quadrant is effectively covered.

Quadratic Equation

Quadratic equations often emerge when analyzing intersection points, and here they played a pivotal role. The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). By applying either the factoring method, completing the square, or the quadratic formula, one can solve for the variable, identifying critical values for curves or lines.
In the given exercise,

• **Appearance:** The equation \( y^2 + y - 6 = 0 \) emerged during the intersection analysis of \(x = 6-y\) and \(y = \sqrt{x}\).

• **Solving quadratics:** Using factoring, this equation becomes \((y+3)(y-2) = 0\), revealing solutions \(y=-3\) and \(y=2\). Since only the first quadrant concerns us, \(y=2\) was the relevant solution.
• **Importance for integration:** These solutions inform the variable bounds and thus influence the domain over which our function is integrated. Quadratic equations frequently offer the constraints or intersections necessary to define realistic integration regions.

Understanding quadratic equations' function aids in processing complex regions in planar geometry successfully, hence advancing deeper comprehension in integral calculus.