Answer to Question #267988 in Calculus for anggggg1

Question #267988

Find the area of the surface that is generated by revolving the portion of the curve y=x^2 between x=0 and x=1 about the y-axis.


1
Expert's answer
2021-11-18T17:54:43-0500

Solution:

The curve y =  x2\ x^{2} from x = 0 and x = 1 about the y axis


We know that x = 0 to x = 1 , which is analogous from y = 0 to y = 1.

Area of the surface is

A=2×πabg(y)1+(g(y)2)dyA = 2\times \pi \int_{a}^{b} g(y)\sqrt{1+(g'(y)^{2})}dy


g(y)=yg(y) = \sqrt{y}

g(y)=1(2×y)g'(y) = \frac {1}{(2\times\sqrt{y})}

A=2×π01y1+(1(2×y))2dyA = 2\times \pi \int_{0}^{1} \sqrt {y} \sqrt{1+(\frac{1}{(2\times\sqrt{y})})^{2}}dy

A=2×π01y(1+14y)dy\Rightarrow A = 2\times \pi \int_{0}^{1} \sqrt{y(1+\frac{1}{4y})}dy

A=2×π01y+(14)dy\Rightarrow A = 2\times \pi \int_{0}^{1} \sqrt{y+(\frac{1}{4})}dy

Let,

u=y+14u = y + \frac {1}{4}


du=dy\Rightarrow du = dy


y=0,u=14y = 0 , u = \frac {1} {4}


y=1,u=54y = 1 , u = \frac {5} {4} ​


A=2×π1454u12du\Rightarrow A = 2\times \pi \int_{\frac{1}{4}}^{\frac{5}{4}} u^ { \frac {1} {2} } du


A=2×π[23(u32)]1454\Rightarrow A= 2 \times \pi [\frac{2}{3} (u^\frac {3} {2}) ]|_{\frac{1}{4}} ^{ \frac{5}{4}}


A=4×π3[(54)32(14)32]\Rightarrow A= \frac{4 \times \pi}{3} [(\frac{5}{4})^\frac {3} {2} - (\frac{1}{4})^\frac {3} {2}]


A=5.33sq.unit\Rightarrow A= 5.33 sq.unit

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