aufgabe 3

a §

b §

$$\begin{array}{rl}A& ={\int }_0^4\sqrt{x}dx-{\int }_0^4\frac{1}{2}xdx\\ & ={\left[\frac{2}{3}{x}^{\frac{3}{2}}\right]}_0^4-{\left[\frac{1}{4}{x}^2\right]}_0^4\\ & =\frac{2}{3}\cdot 4^{\frac{3}{2}}-\frac{1}{4}\cdot 4^2\\ & =\frac{2}{3}\cdot \sqrt{64}-4\\ & =\frac{16}{3}-4\\ & =\underline{1\frac{1}{3}}\end{array}$$

c §

$$\begin{array}{rl}V& =V_f-V_g\\ & =\left(\pi {\int }_0^{4}xdx\right)-\left(\pi {\int }_0^4\frac{1}{4}{x}^2\right)\\ & =\left(\pi {\left[\frac{1}{2}{x}^2\right]}_0^4\right)-\left(\pi {\left[\frac{1}{12}{x}^3\right]}_0^4\right)\\ & =\left(\pi \cdot \frac{1}{2}\cdot 4^2\right)-\left(\pi \cdot \frac{1}{12}{4}^3\right)\\ & =\pi \left(8-\frac{16}{3}\right)\\ & =\underline{\pi \cdot 2\frac{2}{3}}\end{array}$$