Is my work for the following surface area problem using integrals correct?
$begingroup$
I had to compute the surface area of portion of surface $z^2 = 2xy$ which lies above the first quadrant of X-Y planeand is cut off by the planes $x=2$ and $y=1$. Here's my work :
Since,
$$A(S) = int int_{D} sqrt{1 + left(frac{partial z}{partial x}right)^2 + left(frac{partial z}{partial y}right)^2} dA$$
First I took $z = sqrt{2xy}$ and then I took $D$ as $0 leq y leq 1$ and $0 leq x leq 2$ and doubled the result I got.
Is my method correct?
multivariable-calculus surface-integrals
asked Dec 31 '18 at 0:16
AlexAlex
836
$endgroup$
add a comment |
$begingroup$
I had to compute the surface area of portion of surface $z^2 = 2xy$ which lies above the first quadrant of X-Y planeand is cut off by the planes $x=2$ and $y=1$. Here's my work :
Since,
$$A(S) = int int_{D} sqrt{1 + left(frac{partial z}{partial x}right)^2 + left(frac{partial z}{partial y}right)^2} dA$$
First I took $z = sqrt{2xy}$ and then I took $D$ as $0 leq y leq 1$ and $0 leq x leq 2$ and doubled the result I got.
Is my method correct?
multivariable-calculus surface-integrals
asked Dec 31 '18 at 0:16
AlexAlex
836
$endgroup$
add a comment |
$begingroup$
I had to compute the surface area of portion of surface $z^2 = 2xy$ which lies above the first quadrant of X-Y planeand is cut off by the planes $x=2$ and $y=1$. Here's my work :
Since,
$$A(S) = int int_{D} sqrt{1 + left(frac{partial z}{partial x}right)^2 + left(frac{partial z}{partial y}right)^2} dA$$
First I took $z = sqrt{2xy}$ and then I took $D$ as $0 leq y leq 1$ and $0 leq x leq 2$ and doubled the result I got.
Is my method correct?
multivariable-calculus surface-integrals
asked Dec 31 '18 at 0:16
AlexAlex
836
$endgroup$
I had to compute the surface area of portion of surface $z^2 = 2xy$ which lies above the first quadrant of X-Y planeand is cut off by the planes $x=2$ and $y=1$. Here's my work :
Since,
$$A(S) = int int_{D} sqrt{1 + left(frac{partial z}{partial x}right)^2 + left(frac{partial z}{partial y}right)^2} dA$$
First I took $z = sqrt{2xy}$ and then I took $D$ as $0 leq y leq 1$ and $0 leq x leq 2$ and doubled the result I got.
Is my method correct?
multivariable-calculus surface-integrals
multivariable-calculus surface-integrals
asked Dec 31 '18 at 0:16
AlexAlex
836
asked Dec 31 '18 at 0:16
AlexAlex
836
asked Dec 31 '18 at 0:16
AlexAlex
836
asked Dec 31 '18 at 0:16
AlexAlex
836
asked Dec 31 '18 at 0:16
AlexAlex
836
836
add a comment |
add a comment |
1 Answer
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$begingroup$
Your method looks fine, except "above the first quadrant" usually denotes $zgeq 0$ so doubling the result would be unnecessary. So you should get:
$$displaystyle A(S)=int_{0}^{1}int_{0}^{2}dx,dysqrt{1+frac{y}{2x}+frac{x}{2y}}=4$$
answered Dec 31 '18 at 0:40
legendarierslegendariers
737
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Your method looks fine, except "above the first quadrant" usually denotes $zgeq 0$ so doubling the result would be unnecessary. So you should get:
$$displaystyle A(S)=int_{0}^{1}int_{0}^{2}dx,dysqrt{1+frac{y}{2x}+frac{x}{2y}}=4$$
answered Dec 31 '18 at 0:40
legendarierslegendariers
737
$endgroup$
add a comment |
$begingroup$
Your method looks fine, except "above the first quadrant" usually denotes $zgeq 0$ so doubling the result would be unnecessary. So you should get:
$$displaystyle A(S)=int_{0}^{1}int_{0}^{2}dx,dysqrt{1+frac{y}{2x}+frac{x}{2y}}=4$$
answered Dec 31 '18 at 0:40
legendarierslegendariers
737
$endgroup$
add a comment |
$begingroup$
Your method looks fine, except "above the first quadrant" usually denotes $zgeq 0$ so doubling the result would be unnecessary. So you should get:
$$displaystyle A(S)=int_{0}^{1}int_{0}^{2}dx,dysqrt{1+frac{y}{2x}+frac{x}{2y}}=4$$
answered Dec 31 '18 at 0:40
legendarierslegendariers
737
$endgroup$
Your method looks fine, except "above the first quadrant" usually denotes $zgeq 0$ so doubling the result would be unnecessary. So you should get:
$$displaystyle A(S)=int_{0}^{1}int_{0}^{2}dx,dysqrt{1+frac{y}{2x}+frac{x}{2y}}=4$$
answered Dec 31 '18 at 0:40
legendarierslegendariers
737
edited Dec 31 '18 at 0:56
answered Dec 31 '18 at 0:40
legendarierslegendariers
737
answered Dec 31 '18 at 0:40
legendarierslegendariers
737
answered Dec 31 '18 at 0:40
legendarierslegendariers
737
737
add a comment |
add a comment |
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