Find the area of the part of plane $ z = y + 1$ which is inside cylinder $x^2 + y^2 = 1$, $z$ is free

105 Views Asked by At

What I have tried so far :

$$ S = \int\int \sqrt{(0)^2 + (1)^2 + 1} dA$$ Converting to spherical coordinates: $$ \int_{0}^{4\pi}\int_{0}^{4} \sqrt{2} \space r \space drd\theta = \int_{0}^{4\pi} \frac{r^2}{2} f4 \space 0 = \int_{0}^{4\pi} 8 \space d\theta $$ Fairly sure this is incorrect as I dont know where my theta has run off to. What did I do wrong in the integration ?

1

There are 1 best solutions below

0
On BEST ANSWER

Plane would cut the cylinder at 45 degree, thus making an ellipse, making the radius of longer side $\sqrt2$. Since along x axis there is no such constraint, the radius of ellipse remains 1 in that direction. Area of ellipse = $\pi*a*b$ = $\pi*\sqrt2*1$ = $\sqrt2\pi$

\begin{align} Ans = \bbox[yellow,5px,border:2px solid red]{\sqrt2\pi} \end{align}