Area Under Curves Questions (6)

Let $S=\{(x,y)\in\mathbb{R}\times\mathbb{R}: x\geq 0,\,y\geq 0,\,y^2\leq 4x,\,y^2\leq 12-2x,\,3y+8x\leq 58\}$. If the area of $S$ is $\alpha\sqrt{2}$, then $\alpha$ equals:

Let $f(x)=\min\{|x|+1,\,3-|x|\}$ and $g(x)=\sqrt{9-x^2}$. If $S$ is the region in the $xy$-plane defined by $S=\{(x,y): f(x)\leq y\leq g(x)\}$, then the area of $S$ is:

Let $f(x)=\min\{|x|+1,\,3-|x|\}$ and $g(x)=\sqrt{9-x^2}$. If $S$ is the region in the $xy$-plane defined by $S=\{(x,y): f(x)\leq y\leq g(x)\}$, then the area of $S$ is:

If the area bounded by $x = \sqrt{r^2 - y^2}$ and $y = \tan\theta\,(x + r)$ is $\dfrac{1}{2}\pi r^2$, where $r \in \mathbb{R}^+$ and $0 < \theta < \dfrac{\pi}{2}$, then:

Let $S=\{(x,y)\in\mathbb{R}\times\mathbb{R}: x\geq 0,\,y\geq 0,\,y^2\leq 4x,\,y^2\leq 12-2x,\,3y+8x\leq 58\}$. If the area of $S$ is $\alpha\sqrt{2}$, then $\alpha$ equals:

If the area bounded by $x = \sqrt{r^2 - y^2}$ and $y = \tan\theta\,(x + r)$ is $\dfrac{1}{2}\pi r^2$, where $r \in \mathbb{R}^+$ and $0 < \theta < \dfrac{\pi}{2}$, then: