19-4q+q-11 simply the experssion
1 answer:
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We know, that the <span>area of the surface generated by revolving the curve y about the x-axis is given by:
In this case a = 0, b = 15,
and:
So there will be:
![\left(\star\right)=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\dfrac{2\pi}{15}\cdot\dfrac{25}{6}\cdot\left[\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}\right]_0^{15}=\\\\\\= \dfrac{5\pi}{9}\left[\left(1+\dfrac{15^4}{25}\right)^\frac{3}{2}-\left(1+\dfrac{0^4}{25}\right)^\frac{3}{2}\right]=\dfrac{5\pi}{9}\left[2026^\frac{3}{2}-1^\frac{3}{2}\right]=\\\\\\= \boxed{\dfrac{5\Big(2026^\frac{3}{2}-1\Big)}{9}\pi}](https://tex.z-dn.net/?f=%5Cleft%28%5Cstar%5Cright%29%3D%5Cdfrac%7B2%5Cpi%7D%7B15%7D%5Ccdot%5Cint%5Climits_0%5E%7B15%7Dx%5E3%5Ccdot%5Csqrt%7B1%2B%5Cdfrac%7Bx%5E4%7D%7B25%7D%7D%5C%2C%5C%2C%20dx%3D%5Cdfrac%7B2%5Cpi%7D%7B15%7D%5Ccdot%5Cdfrac%7B25%7D%7B6%7D%5Ccdot%5Cleft%5B%5Cleft%281%2B%5Cdfrac%7Bx%5E4%7D%7B25%7D%5Cright%29%5E%5Cfrac%7B3%7D%7B2%7D%5Cright%5D_0%5E%7B15%7D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Cdfrac%7B5%5Cpi%7D%7B9%7D%5Cleft%5B%5Cleft%281%2B%5Cdfrac%7B15%5E4%7D%7B25%7D%5Cright%29%5E%5Cfrac%7B3%7D%7B2%7D-%5Cleft%281%2B%5Cdfrac%7B0%5E4%7D%7B25%7D%5Cright%29%5E%5Cfrac%7B3%7D%7B2%7D%5Cright%5D%3D%5Cdfrac%7B5%5Cpi%7D%7B9%7D%5Cleft%5B2026%5E%5Cfrac%7B3%7D%7B2%7D-1%5E%5Cfrac%7B3%7D%7B2%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Cboxed%7B%5Cdfrac%7B5%5CBig%282026%5E%5Cfrac%7B3%7D%7B2%7D-1%5CBig%29%7D%7B9%7D%5Cpi%7D)
Answer C.
</span>
In this case a = 0, b = 15,
So there will be:
Answer C.
</span>
