Answer: yt
Step-by-step explanation:
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.
Answer:
m<RPS = 53 degrees
Step-by-step explanation:
Find the diagram attached
From the diagram shown;
m<QPR + m<RPS = 90
Given that;
m<QPR = x
m<RPS = x+16
Substitute into the formula;
m<QPR + m<RPS = 90
x+x+16 = 90
2x+16 = 90
2x = 90-16
2x = 74
x = 74/2
x = 37degrees
Get m<RPS;
Since m<RPS = x+16
m<RPS = 37 + 16
m<RPS = 53 degrees
Answer:
the total area is 155 + 23.75 +23.75 = 202.5
Step-by-step explanation:
rectangle = 15.5 times 10 =155
tryangle 1 = 4.75 times 10 divided by 2=23.75
tryangle 2 = 23.75
Answer:
3/4
Step-by-step explanation: