Surface of REvolution

[sareen]

Find the surface area when the region bounded by f(x)=x^2 x=0 x=1 is rotated about the y-axis

My solution:
So, since y=x^2 => x=?y

Is this right?

S=?2?x dx

What's our ds=?1+(dy/dx)^2 dx going to be???

 

[Deleted member 4993]

Find the surface area when the region bounded by f(x)=x^2 x=0 x=1 is rotated about the y-axis

My solution:
So, since y=x^2 => x=?y

Is this right?

S=?2?x dx <<<< I don't know how did you get thatWhat's our ds=?1+(dy/dx)^2 dx going to be???

y = x^2

y' = 2x

\(\displaystyle S = \int_{0}^{1}2\pi x\sqrt{1+4x^2}dx\)

 

[sareen]

ohhhhhhh ok......thanks

 

[BigGlenntheHeavy]

\(\displaystyle Surface \ Area \ around \ y-axis \ is \ S \ = \ \int_{c}^{d}2\pi g(y)\sqrt(1+[g \ ' \ (y)]^{2})dy\)

\(\displaystyle Hence, \ for \ your \ problem, \ x \ = \ g(y) \ = \ 1-y^{1/2}, \ S \ = \ 2\pi\int_{0}^{1}\bigg[(1-y^{1/2})\sqrt(1+\frac{1}{4y})\bigg]dy\)