I am to find the area R bounded by the kurve $x=y^2+2$, $y=x-4$ and $y=0$.
I found the points of intersection by first setting the lines equal eaxh other and used the quadratic formula:
\begin{align*}
y^2+2=y+4
\\y^2-y-2=0
\\
\\y_{1,2}=\frac{(-b) +/- \sqrt{{b^2}-4ac}}{2a}=\frac{-(-1)+/-\sqrt{{(-1)^2}-4*1*(-2)}}{2}=\frac{1 +/- \sqrt{9}}{2}=\frac{1+/-3}{2}\\
\\y_1 = 2 \\y_2=-1
\end{align*}
I am using $y_1=2$ and $y_2=0$ as boundaries because $y=0=x$ is given. Then I sliced horizontally:
\begin{align*}
A=\int_0^2{(y+4)-(y^2+2)}dy\\
A=\int_0^2{2+y-y^2}dy\\
A=[2y-\frac{y^2}{2}-\frac{y^3}{3}]_0^2\\
A=[(2*2)-\frac{2^2}{2}-\frac{2^3}{3}]-[(2*0)-\frac{0^2}{2}-\frac{0^3}{3}]\\
A=4-2-\frac{8}{3}\\
A=-\frac{2}{3}
\end{align*}
I get a negative area, what am I doing wrong here?\
If I slice it vertically, I use the boundaries $x_1=2$ and $x_2=6$ and get the integral
\begin{align*}
A=\int_2^6{(x-4)-(\sqrt{x-2})}dx\\
A=\int_2^6{1-\frac{1}{2\sqrt{x-2}}}\\
A=[1-\frac{1}{2\sqrt{x-2}}]\\
A=(1-\frac{1}{2\sqrt{4}})-(1-\frac{1}{2\sqrt{0}})\\
A=(1-\frac{1}{4})-(1-undefined)
\end{align*}
What is the proper way to solve this?
Best Answer
You made a mistake in the sign when integrating \begin{align*} A=\int_0^2{(y+4)-(y^2+2)}dy\\ A=\int_0^2{2+y-y^2}dy\\ A=[2y\color{red}{+}\frac{y^2}{2}-\frac{y^3}{3}]_0^2\\ A=[(2*2)\color{red}{+}\frac{2^2}{2}-\frac{2^3}{3}]-[(2*0)\color{red}{+}\frac{0^2}{2}-\frac{0^3}{3}]\\ A=4\color{red}{+}2-\frac{8}{3}\\ A=\frac{10}{3} \end{align*}
Integrating wrt to $x$ we have $$A=\int_{2}^{4}\sqrt{x-2}dx+\int_{4}^{6}(\sqrt{x-2}-(x-4))dx$$ $$=\int_{2}^{6}\sqrt{x-2}dx+\int_{4}^{6}(4-x)dx$$ $$=\frac{2}{3}{(x-2)^{\frac{3}{2}}}\big|_{x=2}^{x=6}+\big[4x-\frac{x^2}{2}\big]_{x=4}^{x=6}$$ $$=\frac{16}{3}+\big[6-8]=\frac{16}{3}-2=\frac{10}{3}$$