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Calculate the volume of the solid formed by rotating the area between y=x2y = x^2 ...
Apr 18, 2024
Calculate the volume of the solid formed by rotating the area between y=x2y = x^2 and y=0y = 0 around the x-axis from x=0x = 0 to x=1x = 1.
Generated Graph
Solution by Steps
step 1
To find the volume of the solid formed by rotating the area between y=x2y = x^2 and y=0y = 0 around the x-axis, we use the disk method
step 2
The volume VV of the solid is given by the integral V=∫abπr2dxV = \int_{a}^{b} \pi r^2 dx, where rr is the radius of the disk at a given xx, which is y=x2y = x^2 in this case
step 3
Substituting y=x2y = x^2 into the formula, we get V=∫01π(x2)2dx=∫01πx4dxV = \int_{0}^{1} \pi (x^2)^2 dx = \int_{0}^{1} \pi x^4 dx
step 4
Using the asksia-ll calculator result, we integrate to find the volume: ∫01πx4dx=πx55∣01\int_{0}^{1} \pi x^4 dx = \frac{\pi x^5}{5} \Big|_0^1
step 5
Evaluating the integral from 00 to 11, we get π(1)55−π(0)55=π5\frac{\pi (1)^5}{5} - \frac{\pi (0)^5}{5} = \frac{\pi}{5}
Answer
The volume of the solid is π5\frac{\pi}{5} cubic units.
Key Concept
Disk Method for Volume
Explanation
The disk method involves integrating the area of circular disks along the axis of revolution to find the volume of the solid formed.
Find the Laurent series for the function 1z+z2 in each of the following domains: (a) 0<|z|<1 (b) 1<|z| (c) 0<|z+1|<1 (d) 1<|z+1| ，
Solution by Steps
step 1
To find the Laurent series for the function 1z+z2\frac{1}{z+z^2} around z=0z=0 for 0<|z|<1, we first rewrite the function to separate the terms
step 2
We factor out zz from the denominator to get 1z(1+z)\frac{1}{z(1+z)}
step 3
Next, we perform partial fraction decomposition to get 1z−11+z\frac{1}{z} - \frac{1}{1+z}
step 4
We then expand 11+z\frac{1}{1+z} as a geometric series for |z|<1, which gives 11+z=1−z+z2−z3+…\frac{1}{1+z} = 1 - z + z^2 - z^3 + \ldots
step 5
The Laurent series for 0<|z|<1 is then 1z−(1−z+z2−z3+…)\frac{1}{z} - (1 - z + z^2 - z^3 + \ldots)
Answer
1z−1+z−z2+z3−…\frac{1}{z} - 1 + z - z^2 + z^3 - \ldots
Key Concept
Laurent series expansion in annular regions
Explanation
For 0<|z|<1, we can expand the function as a Laurent series by using partial fraction decomposition and geometric series expansion.
---
step 1
To find the Laurent series for the function 1z+z2\frac{1}{z+z^2} around z=0z=0 for |z|>1, we first rewrite the function to separate the terms
step 2
We factor out z2z^2 from the denominator to get 1z2(1+1z)\frac{1}{z^2(1+\frac{1}{z})}
step 3
Next, we expand 11+1z\frac{1}{1+\frac{1}{z}} as a geometric series for |\frac{1}{z}|<1, which is equivalent to |z|>1
step 4
The geometric series expansion gives 11+1z=1−1z+1z2−1z3+…\frac{1}{1+\frac{1}{z}} = 1 - \frac{1}{z} + \frac{1}{z^2} - \frac{1}{z^3} + \ldots
step 5
The Laurent series for |z|>1 is then 1z2−1z3+1z4−1z5+…\frac{1}{z^2} - \frac{1}{z^3} + \frac{1}{z^4} - \frac{1}{z^5} + \ldots
Answer
1z2−1z3+1z4−1z5+…\frac{1}{z^2} - \frac{1}{z^3} + \frac{1}{z^4} - \frac{1}{z^5} + \ldots
Key Concept
Laurent series expansion outside the unit circle
Explanation
For |z|>1, we expand the function as a Laurent series by factoring out z2z^2 and using a geometric series expansion for the reciprocal of zz.
---
step 1
To find the Laurent series for the function 1z+z2\frac{1}{z+z^2} around z=−1z=-1 for 0<|z+1|<1, we first perform a substitution w=z+1w = z + 1
step 2
The function becomes 1(w−1)+w2\frac{1}{(w-1)+w^2}, which simplifies to 1w2−1\frac{1}{w^2-1}
step 3
We then perform partial fraction decomposition to get 12(1w−1−1w+1)\frac{1}{2}(\frac{1}{w-1} - \frac{1}{w+1})
step 4
Next, we expand 1w−1\frac{1}{w-1} and 1w+1\frac{1}{w+1} as geometric series for |w-1|<1 and |w+1|>1, respectively
step 5
The Laurent series for 0<|z+1|<1 is then 12((−1w−1w2−1w3−…)−(1w−1w2+1w3−…))\frac{1}{2}((-\frac{1}{w} - \frac{1}{w^2} - \frac{1}{w^3} - \ldots) - (\frac{1}{w} - \frac{1}{w^2} + \frac{1}{w^3} - \ldots))
Answer
−1w2−2w4−2w6−…-\frac{1}{w^2} - \frac{2}{w^4} - \frac{2}{w^6} - \ldots where w=z+1w = z + 1
Key Concept
Laurent series expansion around a shifted pole
Explanation
For 0<|z+1|<1, we use the substitution w=z+1w = z + 1 and expand the function as a Laurent series using partial fraction decomposition and geometric series expansion.
---
step 1
To find the Laurent series for the function 1z+z2\frac{1}{z+z^2} around z=−1z=-1 for |z+1|>1, we first perform a substitution w=z+1w = z + 1
step 2
The function becomes 1(w−1)+w2\frac{1}{(w-1)+w^2}, which simplifies to 1w2−1\frac{1}{w^2-1}
step 3
We then perform partial fraction decomposition to get 12(1w−1−1w+1)\frac{1}{2}(\frac{1}{w-1} - \frac{1}{w+1})
step 4
Next, we expand 1w−1\frac{1}{w-1} and 1w+1\frac{1}{w+1} as geometric series for |w-1|>1 and |w+1|<1, respectively
step 5
The Laurent series for |z+1|>1 is then 12((1w+1w2+1w3+…)−(−1w+1w2−1w3+…))\frac{1}{2}((\frac{1}{w} + \frac{1}{w^2} + \frac{1}{w^3} + \ldots) - (-\frac{1}{w} + \frac{1}{w^2} - \frac{1}{w^3} + \ldots))
Answer
1w+2w3+2w5+…\frac{1}{w} + \frac{2}{w^3} + \frac{2}{w^5} + \ldots where w=z+1w = z + 1
Key Concept
Laurent series expansion around a shifted pole outside the unit circle
Explanation
For |z+1|>1, we use the substitution w=z+1w = z + 1 and expand the function as a Laurent series using partial fraction decomposition and geometric series expansion.
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