Answer to Question #282159 in Calculus for james

Question

Find the area bounded by the ff. curves and lines:

1. the loop of y2 = x 4 (3 - x)

2. 4y = x 2 – 2lnx, y = 0, x = 1, x = 4.

3. y 2 + 4x – 8 = 0, Y = 2x

4. an arch of y =1/3sin(x/3)

Answer 1

1.

$$y^2 = x ^4 (3 - x), x\leq3$$

$$y=\pm x^2\sqrt{3-x}$$

$$x^2\sqrt{3-x}=- x^2\sqrt{3-x}$$

$$x_1=0, x_2=3$$

$$Area=A=2\displaystyle\int_{0}^{3}x^2\sqrt{3-x}dx$$

$$\int x^2\sqrt{3-x}dx$$

$$u=3-x, x=3-u, dx=-du$$

$$\int x^2\sqrt{3-x}dx=-\int (3-u)^2\sqrt{u}du$$

$$=-\int(9u^{1/2}-6u^{3/2}+u^{5/2})du$$

$$=-(9(\dfrac{2}{3})u^{3/2}-6(\dfrac{2}{5})u^{5/2}+\dfrac{2}{7}u^{7/2})+C$$

$$=-6(3-x)^{3/2}+\dfrac{12}{5}(3-x)^{5/2}-\dfrac{2}{7}(3-x)^{7/2}+C$$

$$A=2[-6(3-x)^{3/2}+\dfrac{12}{5}(3-x)^{5/2}-\dfrac{2}{7}(3-x)^{7/2}]\begin{matrix} 3 \\ 0 \end{matrix}$$

$$=2(0-(-6(3)^{3/2}+\dfrac{12}{5}(3)^{5/2}-\dfrac{2}{7}(3)^{7/2}))$$

$$=\dfrac{288\sqrt{3}}{35} ({units}^2)$$

2.

$$4y = x^ 2 – 2\ln x, y = 0, x = 1, x = 4.$$

$$y=\dfrac{1}{4}x^2-\dfrac{1}{2}\ln x$$

$$Area=A=\displaystyle\int_{1}^{4}(\dfrac{1}{4}x^2-\dfrac{1}{2}\ln x)dx$$

$$\int\ln xdx=x\ln x-\int x(\dfrac{1}{x})dx=x\ln x-x+C$$

$$A=[\dfrac{1}{12}x^3-\dfrac{1}{2}x\ln x+\dfrac{1}{2}x]\begin{matrix} 4\\ 1 \end{matrix}$$

$$=\dfrac{16}{3}-4\ln 2+2-\dfrac{1}{12}+0-\dfrac{1}{2}$$

$$=\dfrac{27}{4}-4\ln 2\ ({units}^2)$$

3.

$$y=2x=>x=\dfrac{1}{2}y$$

$$y^ 2+4x-8=0=>x=-\dfrac{1}{4}y^2+2$$

$$-\dfrac{1}{4}y^2+2=\dfrac{1}{2}y$$

$$y^2+2y-8=0$$

$$y_1=-4, y_2=2$$

$$Area=A=\displaystyle\int_{-4}^{2}(-\dfrac{1}{4}y^2+2-\dfrac{1}{2}y)dy$$

$$=[-\dfrac{1}{12}y^3+2y-\dfrac{1}{4}y^2]\begin{matrix} 2 \\ -4 \end{matrix}$$

$$=-\dfrac{8}{12}+2(2)-\dfrac{4}{4}-(-\dfrac{-64}{12}+2(-4)-\dfrac{16}{4})$$

$$=9({units}^2)$$

4.

$$y=\dfrac{1}{3}\sin(\dfrac{x}{3})$$

$$y=0=>\dfrac{1}{3}\sin(\dfrac{x}{3})=0$$

$$x=3\pi n, n\in \Z$$

$$x_0=0, x_1=3\pi$$

$$Area=A=\displaystyle\int_{0}^{3\pi}\dfrac{1}{3}\sin(\dfrac{x}{3})dx$$

$$=[-\cos(\dfrac{x}{3})]\begin{matrix} 3\pi \\ 0 \end{matrix}=-(-1-1)=2({units}^2)$$