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Question: What is the area of the region enclosed by the curves:

$$2y = 4\sqrt{x},\quad y = 3,\quad \text{and} \quad 2y + 2x = 6. $$ I have tried calculate all the definite integrals but I am not sure which curve I am supposed to subtract and which one is supposed to come first. And also, I am a little confused because there are three lines. I am also not entirely sure of the bounds.

Any ideas?

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    $\begingroup$ Have you plot a graph? $\endgroup$ – Emilio Novati Feb 27 '18 at 14:42
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$y=2\sqrt{x}$

$y=3$

$y=-x+3$

The intersection point(s) of $y=3$ and $y=2\sqrt{x}$ are:

$x=\left(\dfrac 32\right)^2=\dfrac 94$.

The intersection point(s) of $y=-x+3$ and $y=2\sqrt{x}$ are:

$x^2-6x+9=4x$

$x^2-10x+9=0$

$x=9,1$

$x=9$ is extraneous, so $x=1$.

The area enclosed is the area defined by the difference of $S_1$ and $S_2$, where

$$S_1=\int^\frac 94_0(3-2\sqrt x)dx$$

and $$S_2=\int^1_0(-x+3-2\sqrt x)dx$$

and the area $$I=S_1-S_2$$

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Hint: Split it into two integrals. One where $2y+2x=6$ is under $y=3$ and one where $2y=4\sqrt x$ is under $y=3$. Then find where $2y=4\sqrt x$ and $2y+2x=6$ intersect, and that's the bound between the two integrals.

If you don't want two integrals, you can also integrate with respect to $y$, where the height is the difference between $2y=4\sqrt x$ and $2y+2x=6$ and the bounds are from $2$ to $3$.

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The answers given are more specific, but I wanted to give you a more general answer. When you have multiple lines like this, what they are asking you to do is to think about the shape that it makes. For any particular part of the graph, there will only be two curves that matter. However, to find out which two of the three curves are important, you have to plot them, and then solve for the specific point that it intersects.

You can see the plot here:

http://www.wolframalpha.com/input/?i=plot+2y+%3D+4+sqrt(x)+and+y+%3D+3+and+2y+%2B+2x+%3D+6

Notice that it makes a semi-triangular shape. Therefore, the goal is to find the area of the triangle. The top function for the whole thing is y = 3. The bottom function switches from the linear function to the square root function somewhere in the area of x = 1 (just from looking at the graph). To know exactly, you need to solve for where these are both equal to each other.

Then, you find the area between the two functions (the y = 3 and the linear function) until the switchover, and then, after the switchover happens, you find the area between the other two functions (y = 3 and the square root function).

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enter image description here

As far as the bounding curves can be easily expressed either as $y=f(x)$ or $x=f(y)$, there are more options to calculate the area. For example, we can recognize that while $y$ changes its values from 2 to 3, for every value of $y$, $x$ changes accordingly from $3-y$ to $\tfrac14\,y^2$. Hence

\begin{align} S&=\int_2^3\left( \int_{3-y}^{\tfrac14\,y^2}dx\right)dy \\ &=\int_2^3 \left( x\Big|_{3-y}^{\tfrac14\,y^2}\right)\,dy \\ &=\int_2^3 \left(\tfrac14\,y^2 -3+y\right)\,dy \\ &=(\tfrac1{12}\,y^3 -3y+\tfrac12 y^2)\Big|_2^3 =\frac{13}{12}\approx 1.08333 . \end{align}

It can be also verified that the area of the region is slightly less than the area of triangle $ABC$ with the base $|AC|=\tfrac94$ and height $A_y-B_y=1$, $S_{\triangle ABC}=\tfrac12\cdot\tfrac94\cdot1=\tfrac98=1.125$.

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