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Questions tagged [surface-integrals]

In mathematics, a surface integral is a generalization of multiple integrals to integration over surfaces. It can be thought of as the double integral analog of the line integral.

In mathematics, a surface integral is a generalization of multiple integrals to integration over surfaces. It can be thought of as the double integral analog of the line integral. Given a surface, one may integrate over its scalar fields (that is, functions which return scalars as values), and vector fields (that is, functions which return vectors as values).

Read more on wikipedia's entry Surface integral.

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Double integral/ Area Calculation

The area bounded by the parabola $y²=4ax$ and straight line $x+y=3a$ is....? I just drew the graph and got two intersecting area , one with $+ve$ y-axis and above parabola and another with $+ve$ x-axis and below parabola.. then find the area using…
user334155
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Integral over the face of a tetrahedron

I have the scalar function $$f(x,y,z)=xy+y^2+x+z+1$$ defined over the tetrahedron $K$ with vertices $(x_1,y_1,z_1)=(1,2,3)$, $(x_2,y_2,z_2)=(1,2,4)$, $(x_3,y_3,z_3)=(1,3,5)$ and $(x_4,y_4,z_4)=(-1,0,1)$. I need to calculate the integral of $f$ over…
yemino
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Evaluate the Surface Integral $xyz$ $dS$ where $S$ is the surface defined by $2y=\sqrt{9-x}$, $x>0$, and between $z=0$ and $z=3$.

Evaluate the Surface Integral $xyz$ $dS$ where $S$ is the surface defined by $2y=\sqrt{9-x}$, $x>0$, and between $z=0$ and $z=3$. Don't even know where to start with this question.
Mike
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Surface Integral

Integrate $f=\frac{Y}{X}\sqrt{4Z^2 + 1}$ over the portion of the paraboloid $Z=X^2+Y^2$ that lies above the rectangle with the following limits: $ 1\lt X\lt e , 0\lt y\lt 2 $ in the $X-Y$ plane." I know that you use the formula, but i get stuck…
help123
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Finding the volume of 3 dimensional region under the graph of a function.

Im trying to do the following question but im confused. Let W be the three dimensional region under the graph of the function $f(x,y) = \mathrm{e}^{x^2+y^2}$ and over the region in the $(x,y)$ plane defined by $1\leq x^2+y^2 \leq 2$. I know I have…
user3472448
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Closed surface integral

I've been practicing questions on past exams to prepare for my upcoming one, but they didn't come with solutions and one question has really got me stuck: "Consider $\iint\limits _S \vec F . d \vec S$, where $\vec F =xcos^2(\pi z)\hat i +2ycos(\pi…
user248052
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Surface integral, calculate area of triangle

Show that $$\iint_R\left(x+y\right)\mathrm{d}x\mathrm{d}y=1$$ where $R$ is the triangle with vertices at $\left(0,0\right)$, $\left(1,0\right)$ and $\left(0,2\right)$. Having trouble here with this question, am I supposed to just use polar…
Andrew Cavagnino
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Surface integral problem using spherical coordinates

Hi, for this question I used spherical coordinates. $x=\cos(\theta), y=\sin(\theta)$, $z=t$, $ds=d(\theta)dt$ But when I substitute into the equation I get $12\pi$ instead of $24\pi$(the answer). My limits for the first integration was $0$ to $1$…
Andrew Cavagnino
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Surface integral on an inclined ellipse/Stokes theorem

I have encountered a problem related to Stoke´s Theorem. We are given the intersection between a circular base cylinder that is parallel to the z axis (R3), and a plane that cuts it obliquely, and therefore we end up with an ellipse, that from the…
tuso96
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Question on surface integral. The question uses the normal unit vector instead of just the normal vector , don't understand why.

As the title states. I do not know why the solution used the normal unit vector. I would just use r(u,v) = ui + vj + (-u-v)k and ru x rv = i+j+k to get the result. The question and the solution are attached.
John
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Surface area of elliptic paraboloid

I'm trying to write the surface area of the part of the paraboloid: $$z=(1/2)x^2+(1/2)y^2$$ where $$z\leq a^2/2$$ as double integrals in Cartesian and Cylindrical coordinates. For Cartesian coordinates do I need to parameterise?
Edward
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