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#### renyikouniao

##### Member

- Jun 1, 2013

- 41

x^2+4y^2=1

What is its surface area?

Here's my solution:

I use the equation:

S=integral( upper bound: a lower bound: b ) 2(pi)y*[1+(f'(x))^2]^0.5 dx

Since x^2+4y^2=1

y=[(1-x^2)/4]^0.5

dy/dx=-x/[2(1-x^2)^0.5]

S=integral (upper bound: 0.5 lower bound: -1) 2(pi)[(1-x^2)/4]^0.5 * [1 -x/[2(1-x^2)^0.5]]^0.5

And I have no idea how to evaluate this whole thing...Am I right so far?