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Sauron [17]
3 years ago
12

How can you find the lateral area using the perimeter of the base and the height?

Mathematics
1 answer:
DENIUS [597]3 years ago
3 0
You multiply the perimeter of the base by the height of the prism to get the lateral are or. LA=pB*H Lateral Area=perimeter of Base times Height
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I'm reading this as

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy

with \nabla f=(2xe^{-y},2y-x^2e^{-y}).

The value of the integral will be independent of the path if we can find a function f(x,y) that satisfies the gradient equation above.

You have

\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}

Integrate \dfrac{\partial f}{\partial x} with respect to x. You get

\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx
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Differentiate with respect to y. You get

\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]
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\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy
y^2=g(y)+C
g(y)=y^2+C

So you have

f(x,y)=x^2e^{-y}+y^2+C

The gradient is continuous for all x,y, so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e
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