Re: lilith's question at Yahoo! Anwers regarding finding the area bounded by three curves
Hello lilith,
First let's sketch the area to be found, with the bounded area shaded in red:
View attachment 1301
Now, we may integrate with respect to either $x$ or $y$. I prefer using $y$ as we may use one integral, but we will use both methods, the second as a means of checking our result.
To integrate with respect to $y$, we need to find the lower limit of integration, which is the intersection of the slant line and the parabolic curve. As we will need to know both coordinates for both methods, we will equate them to get the $x$-coordinate:
$$2y=3\sqrt{x}=7-4x$$
Square both sides:
$$9x=49-56x+16x^2$$
Arrange in standard quadratic form:
$$16x^2-65x+49=0$$
Factor:
$$(x-1)(16x-49)=0$$
Discarding the extraneous root, we are left with:
$$x=1\,\therefore\,y=\frac{3}{2}$$
Thus, the bounded area $A$ is given by:
$$A=\int_{\frac{3}{2}}^4 \left(\frac{2y}{3} \right)^2-\frac{7-2y}{4}\,dy=\int_{\frac{3}{2}}^4 \frac{4}{9}y^2+\frac{1}{2}y-\frac{7}{4}\,dy$$
Applying the FTOC, we obtain:
$$A=\left[\frac{4}{27}y^3+\frac{1}{4}y^2-\frac{7}{4}y \right]_{\frac{3}{2}}^4=\left(\frac{256}{27}+4-7 \right)-\left(\frac{1}{2}+\frac{9}{16}-\frac{21}{8} \right)=\frac{175}{27}+\frac{25}{16}=\frac{3475}{432}$$
Now let's integrate with respect to $x$:
$$A=\int_{-\frac{1}{4}}^1 4-\left(\frac{7}{2}-2x \right)\,dx+\int_1^{\frac{64}{9}} 4-\left(\frac{3}{2}\sqrt{x} \right)\,dx=\int_{-\frac{1}{4}}^1 2x+\frac{1}{2}\,dx+\int_1^{\frac{64}{9}} 4-\frac{3}{2}\sqrt{x}\,dx$$
Applying the FTOC, we obtain:
$$A=\left[x^2+\frac{1}{2}x \right]_{-\frac{1}{4}}^1+\left[4x-x^{\frac{3}{2}}\right]_1^{\frac{64}{9}}=\left(\frac{3}{2}+\frac{1}{16} \right)+\left(\frac{256}{27}-3 \right)=\frac{25}{16}+\frac{175}{27}=\frac{3475}{432}$$
