Question answer please 2. (5 points) The base of a solid is the region between y = cos x and the x-axis where -susCross- sections perpendicular to the x-axis are isosceles right triangles with hypotenuse along the base. Find the volume of the solid.
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sdution:-\begin{array}{l}\text { volume }=k \int_{-\frac{\lambda}{2}}^{\lambda / 2} y^{2} d x \\=R \int_{-\lambda / 2}^{\lambda^{2}} \cos ^{2} x d x \\=2 \pi \int_{0}^{\frac{1}{2}} \cos ^{2} x d x \cdot\left(\cos ^{2} x \text { is even fimation }\right) \\=\frac{2 \lambda}{x} \int_{0}^{\frac{1}{2}}(\cos 2 x+1) d x \\=\alpha\left[\int_{0}^{\phi / 2} \cos 2 x d x+\int_{0}^{\frac{1}{2}} d x\right] \\=K\left[\left.\frac{\sin 2 x}{2}\right|_{0} ^{\frac{1}{2}}+x \int_{0}^{k_{/ 2}}\right] \\=\frac{1}{2}\left(\sin ^{70} \alpha-\sin 0\right)+\lambda\left(\lambda_{2}-0\right) \\=+\frac{h^{2}}{2}=+\frac{k^{2}}{2} \\\end{array}Hencl Requined volume =\frac{r^{2}}{2} unit' ...