How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$ ? I can't even figure out what the integrand will be ( should it be $\sqrt{z^2-x^2}$ ?
) and not even the limits . Please help . thanks in advance
[Math] How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$
multiple integralmultivariable-calculussurface-integrals
Best Answer
Comment:
The apex of the cone is $(0,0,0)$ and at height $z=t$ the cone admits a representation by a circle with radius $t$ and polar coordinates: $(t\cos \varphi,t\sin \varphi)$.
The limits of the parameter representation $t$ in (2) are due to the planes in (1) \begin{align*} 0&\leq t\leq \frac{3-x}{2}\qquad\text{and}\qquad x=t\cos\varphi \end{align*}
The integral (3) was calculated with the help of WolframAlpha.
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