Consider the graphs of j(x)== sin (6x- pi 2 )+11 and k(x) = 7sin(6x - pi/2) - 11 Which feature of the graph of k(x) is different
1 answer:
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First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:
And the derivative of x:
Now, we can calculate the area of the surface:
We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:
Calculate indefinite integral:
And the area:
![A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}](https://tex.z-dn.net/?f=A%3D2%5Cpi%5Cint%5Climits_0%5E%7B10%7Dx%5Csqrt%7B4x%5E2%2B1%7D%5C%2Cdx%3D2%5Cpi%5Ccdot%5Cdfrac%7B1%7D%7B12%7D%5Cbigg%5B%5Cleft%284x%5E2%2B1%5Cright%29%5E%5Cfrac%7B3%7D%7B2%7D%5Cbigg%5D_0%5E%7B10%7D%3D%5C%5C%5C%5C%5C%5C%3D%20%5Cdfrac%7B%5Cpi%7D%7B6%7D%5Cleft%5B%5Cbig%284%5Ccdot10%5E2%2B1%5Cbig%29%5E%5Cfrac%7B3%7D%7B2%7D-%5Cbig%284%5Ccdot0%5E2%2B1%5Cbig%29%5E%5Cfrac%7B3%7D%7B2%7D%5Cright%5D%3D%5Cdfrac%7B%5Cpi%7D%7B6%7D%5CBig%28%5Cbig401%5E%5Cfrac%7B3%7D%7B2%7D-1%5E%5Cfrac%7B3%7D%7B2%7D%5CBig%29%3D%5Cboxed%7B%5Cdfrac%7B401%5E%5Cfrac%7B3%7D%7B2%7D-1%7D%7B6%7D%5Cpi%7D)
Answer D.
And the derivative of x:
Now, we can calculate the area of the surface:
We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:
Calculate indefinite integral:
And the area:
Answer D.