1. surface integral

By using the double integral , find surface area of the portion of the surface z = sqrt (4 -x^2) that lies above rectangle in xy -plane whose coordinate dsatisfy 0<x<1 and 0<y<4 ...

I have sketched out the diagram , but i found that , it's merely 2 plane surfaces , it's not a 3d object formed by 2 surfaces , right ? So, we just need to find the respective surface area and add 2 things up ?

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2. Re: surface integral

Hey xl5899.

For this object you will need to let y vary and when you sketch it you should get a cylinder like object where the cross section circle is in the x-z plane instead of the x-y plane [where it usually is for a normal cylinder].

Try sketching that to see what it looks like [set y = 0 for the first circle and then draw the same circle across the y-axis to get the cylinder like surface you need to evaluate].

3. Re: surface integral Originally Posted by chiro Hey xl5899.

For this object you will need to let y vary and when you sketch it you should get a cylinder like object where the cross section circle is in the x-z plane instead of the x-y plane [where it usually is for a normal cylinder].

Try sketching that to see what it looks like [set y = 0 for the first circle and then draw the same circle across the y-axis to get the cylinder like surface you need to evaluate].
do you mean copy the same set of xz plane i have now to y = 4 and join the lines between them to form 3d object ?

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