Surface Area of Multiple Integration
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Find the area of the upper half of the cone $x^2+y^2=z^2$ above the interior of one loop of $r=cos(2theta)$.
I know the formula for surface area is $int_{x_0}^{x_1}int_{y_0}^{y_1}sqrt{(f_x)^2+(f_y)^2+1}dxdy$, and in this question it should be $int_{x_0}^{x_1}int_{y_0}^{y_1}sqrt{(frac{-2x}{2sqrt{-y^2-x^2}})^2+(frac{-2y}{2sqrt{-y^2-x^2}})^2+1}dxdy$, but I am not sure about the bounds. I am guessing that since I'm only taking the area above one loop of $r=cos(2theta)$, that the bound for $x$ goes from $0$ to $1$. What about the bounds for $y$? Thanks.
multivariable-calculus
asked Dec 9 '18 at 0:19
pecopeco
888
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add a comment |
$begingroup$
Find the area of the upper half of the cone $x^2+y^2=z^2$ above the interior of one loop of $r=cos(2theta)$.
I know the formula for surface area is $int_{x_0}^{x_1}int_{y_0}^{y_1}sqrt{(f_x)^2+(f_y)^2+1}dxdy$, and in this question it should be $int_{x_0}^{x_1}int_{y_0}^{y_1}sqrt{(frac{-2x}{2sqrt{-y^2-x^2}})^2+(frac{-2y}{2sqrt{-y^2-x^2}})^2+1}dxdy$, but I am not sure about the bounds. I am guessing that since I'm only taking the area above one loop of $r=cos(2theta)$, that the bound for $x$ goes from $0$ to $1$. What about the bounds for $y$? Thanks.
multivariable-calculus
asked Dec 9 '18 at 0:19
pecopeco
888
$endgroup$
$begingroup$
Hint: Expand and simplify the expression under the root sign, after correcting the negative signs of $y^2$ and $x^2$.
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– random
Dec 9 '18 at 0:32
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@random I know how to do the actual integral, but I'm confused as to what the bounds should be.
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– peco
Dec 9 '18 at 0:56
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Doing everything in polar coordinates is an option.
$endgroup$
– random
Dec 9 '18 at 1:45
add a comment |
$begingroup$
Find the area of the upper half of the cone $x^2+y^2=z^2$ above the interior of one loop of $r=cos(2theta)$.
I know the formula for surface area is $int_{x_0}^{x_1}int_{y_0}^{y_1}sqrt{(f_x)^2+(f_y)^2+1}dxdy$, and in this question it should be $int_{x_0}^{x_1}int_{y_0}^{y_1}sqrt{(frac{-2x}{2sqrt{-y^2-x^2}})^2+(frac{-2y}{2sqrt{-y^2-x^2}})^2+1}dxdy$, but I am not sure about the bounds. I am guessing that since I'm only taking the area above one loop of $r=cos(2theta)$, that the bound for $x$ goes from $0$ to $1$. What about the bounds for $y$? Thanks.
multivariable-calculus
asked Dec 9 '18 at 0:19
pecopeco
888
$endgroup$
Find the area of the upper half of the cone $x^2+y^2=z^2$ above the interior of one loop of $r=cos(2theta)$.
I know the formula for surface area is $int_{x_0}^{x_1}int_{y_0}^{y_1}sqrt{(f_x)^2+(f_y)^2+1}dxdy$, and in this question it should be $int_{x_0}^{x_1}int_{y_0}^{y_1}sqrt{(frac{-2x}{2sqrt{-y^2-x^2}})^2+(frac{-2y}{2sqrt{-y^2-x^2}})^2+1}dxdy$, but I am not sure about the bounds. I am guessing that since I'm only taking the area above one loop of $r=cos(2theta)$, that the bound for $x$ goes from $0$ to $1$. What about the bounds for $y$? Thanks.
multivariable-calculus
multivariable-calculus
asked Dec 9 '18 at 0:19
pecopeco
888
asked Dec 9 '18 at 0:19
pecopeco
888
asked Dec 9 '18 at 0:19
pecopeco
888
asked Dec 9 '18 at 0:19
pecopeco
888
asked Dec 9 '18 at 0:19
pecopeco
888
888
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Hint: Expand and simplify the expression under the root sign, after correcting the negative signs of $y^2$ and $x^2$.
$endgroup$
– random
Dec 9 '18 at 0:32
$begingroup$
@random I know how to do the actual integral, but I'm confused as to what the bounds should be.
$endgroup$
– peco
Dec 9 '18 at 0:56
$begingroup$
Doing everything in polar coordinates is an option.
$endgroup$
– random
Dec 9 '18 at 1:45
add a comment |
$begingroup$
Hint: Expand and simplify the expression under the root sign, after correcting the negative signs of $y^2$ and $x^2$.
$endgroup$
– random
Dec 9 '18 at 0:32
$begingroup$
@random I know how to do the actual integral, but I'm confused as to what the bounds should be.
$endgroup$
– peco
Dec 9 '18 at 0:56
$begingroup$
Doing everything in polar coordinates is an option.
$endgroup$
– random
Dec 9 '18 at 1:45
$begingroup$
Hint: Expand and simplify the expression under the root sign, after correcting the negative signs of $y^2$ and $x^2$.
$endgroup$
– random
Dec 9 '18 at 0:32
$begingroup$
Hint: Expand and simplify the expression under the root sign, after correcting the negative signs of $y^2$ and $x^2$.
$endgroup$
– random
Dec 9 '18 at 0:32
$begingroup$
@random I know how to do the actual integral, but I'm confused as to what the bounds should be.
$endgroup$
– peco
Dec 9 '18 at 0:56
$begingroup$
@random I know how to do the actual integral, but I'm confused as to what the bounds should be.
$endgroup$
– peco
Dec 9 '18 at 0:56
$begingroup$
Doing everything in polar coordinates is an option.
$endgroup$
– random
Dec 9 '18 at 1:45
$begingroup$
Doing everything in polar coordinates is an option.
$endgroup$
– random
Dec 9 '18 at 1:45
add a comment |
1 Answer
1
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In cylindrical coordinates it gets simpler. The surface element for this cone with the parametrization $vec s=(rcostheta,rsintheta,r)$ is $mathbb dS=sqrt{2}r,mathbb dr,mathbb dtheta$. So
$$S=dfrac14int_0^{2pi}int_0^{cos(2theta)}sqrt2r,mathbb dr,mathbb dtheta$$
The factor 1/4 is because the graph of the curve has four equal lobes.
answered Dec 9 '18 at 12:14
Rafa BudríaRafa Budría
5,9101825
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In cylindrical coordinates it gets simpler. The surface element for this cone with the parametrization $vec s=(rcostheta,rsintheta,r)$ is $mathbb dS=sqrt{2}r,mathbb dr,mathbb dtheta$. So
$$S=dfrac14int_0^{2pi}int_0^{cos(2theta)}sqrt2r,mathbb dr,mathbb dtheta$$
The factor 1/4 is because the graph of the curve has four equal lobes.
answered Dec 9 '18 at 12:14
Rafa BudríaRafa Budría
5,9101825
$endgroup$
add a comment |
$begingroup$
In cylindrical coordinates it gets simpler. The surface element for this cone with the parametrization $vec s=(rcostheta,rsintheta,r)$ is $mathbb dS=sqrt{2}r,mathbb dr,mathbb dtheta$. So
$$S=dfrac14int_0^{2pi}int_0^{cos(2theta)}sqrt2r,mathbb dr,mathbb dtheta$$
The factor 1/4 is because the graph of the curve has four equal lobes.
answered Dec 9 '18 at 12:14
Rafa BudríaRafa Budría
5,9101825
$endgroup$
add a comment |
$begingroup$
In cylindrical coordinates it gets simpler. The surface element for this cone with the parametrization $vec s=(rcostheta,rsintheta,r)$ is $mathbb dS=sqrt{2}r,mathbb dr,mathbb dtheta$. So
$$S=dfrac14int_0^{2pi}int_0^{cos(2theta)}sqrt2r,mathbb dr,mathbb dtheta$$
The factor 1/4 is because the graph of the curve has four equal lobes.
answered Dec 9 '18 at 12:14
Rafa BudríaRafa Budría
5,9101825
$endgroup$
In cylindrical coordinates it gets simpler. The surface element for this cone with the parametrization $vec s=(rcostheta,rsintheta,r)$ is $mathbb dS=sqrt{2}r,mathbb dr,mathbb dtheta$. So
$$S=dfrac14int_0^{2pi}int_0^{cos(2theta)}sqrt2r,mathbb dr,mathbb dtheta$$
The factor 1/4 is because the graph of the curve has four equal lobes.
answered Dec 9 '18 at 12:14
Rafa BudríaRafa Budría
5,9101825
edited Dec 10 '18 at 5:36
answered Dec 9 '18 at 12:14
Rafa BudríaRafa Budría
5,9101825
answered Dec 9 '18 at 12:14
Rafa BudríaRafa Budría
5,9101825
answered Dec 9 '18 at 12:14
Rafa BudríaRafa Budría
5,9101825
5,9101825
add a comment |
add a comment |
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$begingroup$
Hint: Expand and simplify the expression under the root sign, after correcting the negative signs of $y^2$ and $x^2$.
$endgroup$
– random
Dec 9 '18 at 0:32
$begingroup$
@random I know how to do the actual integral, but I'm confused as to what the bounds should be.
$endgroup$
– peco
Dec 9 '18 at 0:56
$begingroup$
Doing everything in polar coordinates is an option.
$endgroup$
– random
Dec 9 '18 at 1:45