The arc of the parabola y = x^2 from (3,9) to (4,16) is rotated about the y-axis. Find the area of the resulting surface. Please
1 answer:
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Answer:
31.4 in
Step-by-step explanation:
the formula for circumference is 2**radius. Since the diameter (which is the radius multiplied by 2) is labeled we can divide 10 by 2 to get the radius. 10/2 is 5. Since pi is an infinite number, we round it to 3.14. Then, plug in the numbers. The expression will be 2*3.14*5. 2* = 10 and 10 * 3.14 you can just move the decimal so the answer is 31.4.
Check the picture below.
so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).
because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,
![\bf \begin{cases} h=2\\ k=-3\\ a=6\\ c=2\sqrt{10} \end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1 \\\\\\ \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1 \\\\\\ c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2 \\\\\\ 4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2 \\\\\\ \sqrt{4}=b\implies 2=b\\\\ -------------------------------\\\\ \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ah%3D2%5C%5C%0Ak%3D-3%5C%5C%0Aa%3D6%5C%5C%0Ac%3D2%5Csqrt%7B10%7D%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B%5By-%20%28-3%29%5D%5E2%7D%7B%206%5E2%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%20b%5E2%7D%3D1%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%28y%2B3%29%5E2%7D%7B%2036%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%20b%5E2%7D%3D1%0A%5C%5C%5C%5C%5C%5C%0Ac%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%282%5Csqrt%7B10%7D%29%5E2%3D6%5E2%2Bb%5E2%5Cimplies%202%5E2%28%5Csqrt%7B10%7D%29%5E2%3D36%2Bb%5E2%0A%5C%5C%5C%5C%5C%5C%0A4%2810%29%3D36%2Bb%5E2%5Cimplies%2040%3D36%2Bb%5E2%5Cimplies%204%3Db%5E2%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%7B4%7D%3Db%5Cimplies%202%3Db%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ccfrac%7B%28y%2B3%29%5E2%7D%7B%2036%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%202%5E2%7D%3D1%5Cimplies%20%5Ccfrac%7B%28y%2B3%29%5E2%7D%7B%2036%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%204%7D%3D1)
so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).
because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,