Find the area of the surface generated when the given curve is revolved about the given axis. 5x^1/3
1 answer:
Answer:
![\mathbf{ \dfrac{\pi}{675}\Big[ 34\sqrt{34} -125\Big] }](https://tex.z-dn.net/?f=%5Cmathbf%7B%20%5Cdfrac%7B%5Cpi%7D%7B675%7D%5CBig%5B%2034%5Csqrt%7B34%7D%20-125%5CBig%5D%20%7D)
Step-by-step explanation:
The curve x = f(y)
The area of the surface around the y-axis from y = a → y = b is:

From the given curve:
; assuming the region bounded by the curve is 0 ≤ y ≤ 1
So;

5x = y³

The differential of the above equation Is:


Now, we have the area of the surface produced around the curve
through the y axis from the region y = 0 to y = 1
∴





Let make 
It implies that:




when y = 0 ;


u = 5
when y = 1;



∴
The equation
can be written as:



![= \dfrac{\pi}{225}\Big[ \dfrac{u^3}{3} \Big] ^{\sqrt{34}}_{3}\\](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B%5Cpi%7D%7B225%7D%5CBig%5B%20%5Cdfrac%7Bu%5E3%7D%7B3%7D%20%5CBig%5D%20%5E%7B%5Csqrt%7B34%7D%7D_%7B3%7D%5C%5C)
![\mathbf{= \dfrac{\pi}{675}\Big[ 34\sqrt{34} -125\Big] }](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D%20%5Cdfrac%7B%5Cpi%7D%7B675%7D%5CBig%5B%2034%5Csqrt%7B34%7D%20-125%5CBig%5D%20%7D)
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