Answer to Question #342412 in Calculus for Jane

Question

Find the area, take the elements of the area perpendicular to the x-axis. x²-y+1=0; x-y+1=0.

Answer 1

$$x^2-y+1=0=>y=x^2+1$$

$$x-y+1=0=>y=x+1$$

$$x^2+1=x+1$$

$$x_1=0, x_2=1$$

$$y(0)=0+1=1$$

$$y(1)=1+1=2$$

$$x^2-y+1=0, x\ge0$$

$$x=\sqrt{y-1}$$

$$x-y+1=0=>x=y-1$$

$$A=\displaystyle\int_{0}^{1}(x+1(-x^2+1))dx$$ $$=\bigg[\dfrac{x^2}{2}-\dfrac{x^3}{3}\bigg]\begin{matrix} 1 \\ 0 \end{matrix}$$ $$=\dfrac{1^2}{2}-\dfrac{1^3}{3}-0=\dfrac{1}{6}({units}^2)$$

$$-(\dfrac{2(1-1)^{3/2}}{3}-\dfrac{(1)^2}{2}+1)=\dfrac{1}{6}({units}^2)$$

$$A=\displaystyle\int_{1}^{2}(\sqrt{y-1}-(y-1))dy$$ $$=\bigg[\dfrac{2(y-1)^{3/2}}{3}-\dfrac{y^2}{2}+y\bigg]\begin{matrix} 2 \\ 1 \end{matrix}$$ $$=\dfrac{2(2-1)^{3/2}}{3}-\dfrac{(2)^2}{2}+2$$

$$-(\dfrac{2(1-1)^{3/2}}{3}-\dfrac{(1)^2}{2}+1)=\dfrac{1}{6}({units}^2)$$