- #1
DryRun
Gold Member
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Evaluate surface area of top hemisphere
Homework Statement
http://s1.ipicture.ru/uploads/20120106/jrHc5122.jpg
The attempt at a solution
This problem is new to me, in the sense that the integration is to be done against S, which is the same region S, over which the limits are defined. Usually, the region and the d.w.r.to is different. For example, it is typical to use D as the underset and then dA instead of dS. dA is to find the area, but i would assume that dS means the same?
The projection is done on the xy-plane so z = g(x,y) is the equation of the surface.
So, making z the subject of formula:
z=\sqrt{1-x^2-y^2}
f_x=\frac{\partial z}{\partial x}=\frac{-x}{\sqrt{1-x^2-y^2}}
f_y=\frac{\partial z}{\partial x}=\frac{-y}{\sqrt{1-x^2-y^2}}
SA=\underset{S}{{\iint}} \sqrt{1+(f_x)^2+(f_y)^2}\: dA
\sqrt{1+(f_x)^2+(f_y)^2} = \frac{1}{\sqrt{1-x^2-y^2}}
The boundaries for circular region S:
For x fixed, y varies from y=0 to y=√(1−x^2)
x varies from x=0 to x=1
Since it's a circular region, i should convert to polar coordinates for easier integration:
For θ fixed, r varies from 0 to 1.
θ varies from 0 to 2∏
The integrand is converted to polar coordinates form:
\frac{1}{\sqrt{1-x^2-y^2}}=\frac{1}{\sqrt{1-r^2}}
Surface\; area\;of\;S=\int_0^{2\pi}\int_0^1 \frac{1}{\sqrt{1-r^2}}\,.rdrd\theta
For some reason, my final integral above is wrong, as I'm supposed to get this instead:
Surface\; area=\int_0^{2\pi}\int_0^1 r^2\,.rdrd\theta
I just followed the standard procedure for finding the surface area, but there is the integrand from the original equation itself that i have not included in that integral, (x^2+y^2)z and i have no idea how.
I'm going to try and convert it to polar coordinates as well:
(x^2+y^2)z=r^2\sqrt{1-r^2}
By instinct, I'm just going to multiply this converted integrand with the other integrand, which gives me r^2 and this gives me the correct answer! Was that a fluke or did i follow the right steps?
Homework Statement
http://s1.ipicture.ru/uploads/20120106/jrHc5122.jpg
The attempt at a solution
This problem is new to me, in the sense that the integration is to be done against S, which is the same region S, over which the limits are defined. Usually, the region and the d.w.r.to is different. For example, it is typical to use D as the underset and then dA instead of dS. dA is to find the area, but i would assume that dS means the same?
The projection is done on the xy-plane so z = g(x,y) is the equation of the surface.
So, making z the subject of formula:
z=\sqrt{1-x^2-y^2}
f_x=\frac{\partial z}{\partial x}=\frac{-x}{\sqrt{1-x^2-y^2}}
f_y=\frac{\partial z}{\partial x}=\frac{-y}{\sqrt{1-x^2-y^2}}
SA=\underset{S}{{\iint}} \sqrt{1+(f_x)^2+(f_y)^2}\: dA
\sqrt{1+(f_x)^2+(f_y)^2} = \frac{1}{\sqrt{1-x^2-y^2}}
The boundaries for circular region S:
For x fixed, y varies from y=0 to y=√(1−x^2)
x varies from x=0 to x=1
Since it's a circular region, i should convert to polar coordinates for easier integration:
For θ fixed, r varies from 0 to 1.
θ varies from 0 to 2∏
The integrand is converted to polar coordinates form:
\frac{1}{\sqrt{1-x^2-y^2}}=\frac{1}{\sqrt{1-r^2}}
Surface\; area\;of\;S=\int_0^{2\pi}\int_0^1 \frac{1}{\sqrt{1-r^2}}\,.rdrd\theta
For some reason, my final integral above is wrong, as I'm supposed to get this instead:
Surface\; area=\int_0^{2\pi}\int_0^1 r^2\,.rdrd\theta
I just followed the standard procedure for finding the surface area, but there is the integrand from the original equation itself that i have not included in that integral, (x^2+y^2)z and i have no idea how.
I'm going to try and convert it to polar coordinates as well:
(x^2+y^2)z=r^2\sqrt{1-r^2}
By instinct, I'm just going to multiply this converted integrand with the other integrand, which gives me r^2 and this gives me the correct answer! Was that a fluke or did i follow the right steps?
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