在抛物线y=x2(x>0)上求一点P,使该抛物线与其在点P处的切线及x轴所围平面图形的面积为,并求该平面图形绕x轴旋转一周所形成的旋转体的体积.正确答案:

步骤 1：确定点P的坐标
设点P的坐标为$(x_0, x_0^2)$，则该点处的切线斜率为$y' = 2x_0$。因此，切线方程为$y - x_0^2 = 2x_0(x - x_0)$，即$y = 2x_0x - x_0^2$。

步骤 2：计算围成的面积
围成的面积由抛物线$y = x^2$、切线$y = 2x_0x - x_0^2$和x轴围成。面积$A$可以通过积分计算得到：
$$A = \int_{0}^{x_0} (x^2 - (2x_0x - x_0^2)) dx = \int_{0}^{x_0} (x^2 - 2x_0x + x_0^2) dx$$
$$= \left[\frac{x^3}{3} - x_0x^2 + x_0^2x\right]_{0}^{x_0} = \frac{x_0^3}{3} - x_0^3 + x_0^3 = \frac{x_0^3}{3}$$
根据题意，$A = \frac{2}{3}$，所以$\frac{x_0^3}{3} = \frac{2}{3}$，解得$x_0 = \sqrt[3]{2}$。

步骤 3：计算旋转体的体积
旋转体的体积$V$可以通过旋转体的体积公式计算得到：
$$V = \pi \int_{0}^{x_0} (x^2)^2 dx - \pi \int_{0}^{x_0} (2x_0x - x_0^2)^2 dx$$
$$= \pi \int_{0}^{x_0} x^4 dx - \pi \int_{0}^{x_0} (4x_0^2x^2 - 4x_0^3x + x_0^4) dx$$
$$= \pi \left[\frac{x^5}{5}\right]_{0}^{x_0} - \pi \left[\frac{4x_0^2x^3}{3} - 2x_0^3x^2 + x_0^4x\right]_{0}^{x_0}$$
$$= \pi \left(\frac{x_0^5}{5} - \frac{4x_0^5}{3} + 2x_0^5 - x_0^5\right) = \pi \left(\frac{x_0^5}{5} - \frac{4x_0^5}{3} + x_0^5\right)$$
$$= \pi \left(\frac{3x_0^5 - 20x_0^5 + 15x_0^5}{15}\right) = \pi \left(\frac{8x_0^5}{15}\right)$$
将$x_0 = \sqrt[3]{2}$代入，得到$V = \frac{8\pi(\sqrt[3]{2})^5}{15} = \frac{8\pi\sqrt[3]{32}}{15} = \frac{16\pi}{15}$。