Review 5: Areas Between Curves - Now You Try I-2

In the region $x\le 1$, the first two curves intersect at $(-2,0)$ and $(1,\sqrt{3})$. When $x=0$ we see that the first curve is at $y=\sqrt{2}$ and the second is at $y=0$, so the second curve is higher than the first in between $x=-2$ and $x=1$. The area is then$\begin{align*} A&=\int_{-2}^1 \sqrt{x+2}-x\sqrt{4-x^2} \, dx \\ &=\tfrac23 (x+2)^{3/2}+\tfrac13 (4-x^2)^{3/2}\Big|_{-2}^1 \\ &=\tfrac23 (3)^{3/2}+\tfrac13 (3)^{3/2} \\ &=3^{3/2}. \end{align*}$

A diagram of the region in question is given below. We can see that the curves $y=1/x$ and $y=\sqrt{2-x^2}$ intersect at the point $(1,1)$, and $y=\sqrt{2-x^2}$ intersects the  $x$-axis at$(\sqrt{2},0)$. Hence we can write the area as an integral $A=\int_1^{\sqrt{2}} \frac1x-\sqrt{2-x^2} \, dx + \int_{\sqrt{2}}^4 \frac1x \, dx$There are two ways to compute this integral, either using trigonometric substitution or geometry.

Approach 1) Let's rewrite $A$ as $A=\int_1^4 \tfrac1x \, dx - \int_1^{\sqrt{2}} \sqrt{2-x^2} \, dx=\ln 4-\int_1^{\sqrt{2}} \sqrt{2-x^2} \, dx.$ To compute the second integral, we use trigonometric substitution; specifically, substitute $x=\sqrt2\sin(\theta)$, so $dx=\sqrt2\cos(\theta) \, d\theta$. Then$\begin{align*} \int_1^{\sqrt{2}} \sqrt{2-x^2} \, dx&=\int_{x=1}^{x=\sqrt{2}} \sqrt{2-2\sin^2(\theta)} \sqrt{2}\cos(\theta) \, d\theta \\ &=\int_{x=1}^{x=\sqrt{2}} 2\cos^2(\theta) \, d\theta \\ &=\int_{x=1}^{x=\sqrt{2}} \cos(2\theta)+1 \, d\theta \\ &=\tfrac12 \sin(2\theta)+\theta\Big|_{x=1}^{x=\sqrt{2}} \\ &=\sin(\theta)\cos(\theta)+\theta\Big|_{x=1}^{x=\sqrt{2}} \\ &=\sin(\theta)\sqrt{1-\sin^2(\theta)}+\theta\Big|_{x=1}^{x=\sqrt{2}} \\ &=\tfrac{x}{\sqrt{2}}\sqrt{1-\tfrac{x^2}{2}}+\arcsin(\tfrac{x}{\sqrt{2}}) \Big|_{1}^{\sqrt{2}} \\ &=0+\arcsin(1)-\tfrac{1}{\sqrt{2}}\sqrt{\tfrac{1}{2}}-\arcsin(\tfrac1{\sqrt{2}}) \\ &=\tfrac\pi4-\tfrac12. \end{align*}$Thus $A=\ln 4+\tfrac12-\pi/4$.

Approach 2) We can use geometry to find the area of the region. The blue curve above is the graph of a semicircle with radius $\sqrt{2}$,  so we can reinterpret the area as the difference between two quantities, the area under the curve $y=1/x$ from $x=1$ to $x=4$, and this red region of the circle This red region is the difference between a sector of the circle and a triangle, as seen below. Thus the area is given by $\begin{align*} A&=\int_1^4 \frac1x \, dx-(Sector-Triangle) \\&=\ln(4)-\ln(1)-\Big(\tfrac{\pi (\sqrt{2})^2}{8}-\tfrac12\Big) \\ &=\ln(4)-\tfrac{\pi}{4}+\tfrac12. \end{align*}$