multivariable-calculussurface-integrals
Find the area of the surface. The part of the plane $5x+2y+z=10$ that lies inside the cylinder $x^2 + y^2 = 25$.
I know the double integral formula for finding surface area and how to find the partial derivatives that go into the formula. But how do I find the limits of integration? I'm stuck on finding the intersection of these two curves.
$z=1-x-y$ so set $f(x,y)=1-x-y$ $$S=\int\int_D\sqrt{f_x^2+f_y^2+1}dxdy$$
$$S=\int\int_D\sqrt{3}dxdy$$
$$S=\sqrt{3}\int\int_Ddxdy$$
$$S=\sqrt{3}(4\pi) $$
$$S=4\sqrt{3}\pi$$
Note: $D$ is the region enclosed by the circle $x^2+y^2=4$ in $x,y$ plane.
The intersection of a cylinder with a plane is an ellipse. Find the semiaxes of the ellipse and you get $$S=\pi ab$$
The minor semiaxis is always the same as the radius of the cylinder, in this case $b=r=2$.
The major semiaxis can be calculated from the angle between the plane and the cylinder axis. The angle the plane makes with the $z$ axis can be extracted from the plane normal, as the dot product gives us a cosine between two vectors, $\cos\alpha=(n_x,n_y,n_z)\cdot(0,0,1)=n_z$.
The normal of your plane can be read directly from the coefficient of the equation. An equation for a plane can be written as a dot product $\vec{n}\cdot\vec{r}=\rm const$, in your case $(1,2,1)\cdot(x,y,z)=4$. Rescale the normal to unit size and you get: $$\vec{n}=\frac{(1,2,1)}{\sqrt 6}$$ and as we demonstrated above, also by using the dot product, the $z$ component of the normal equals the cosine of the angle with the $z$ axis: $$\cos\alpha=\frac{1}{\sqrt 6}$$
If you draw the vertical cross section (the figure on the right), you can see a right triangle that relates the radius of the cylinder with the hypotenuse (the semiaxis):
$$a=\frac{r}{\cos\alpha}=2\sqrt 6$$ leading to the solution $$S=4\pi \sqrt 6$$
EDIT:
Best Answer
The surface can be parametrized as follows: $$ x=x,\quad y=y, \quad z=10-5x-2y\quad (x,y)\in D $$ with $$D=\{(x,y)\;|\; x^2+y^2\le 25 \} $$ It follows that the wanted area equals $$ A=\iint_D ||r_x\times r_y || \;dA =\iint_D \sqrt{5^2+2^2+1^1} \;dA =\sqrt{30}\;A(D)=\sqrt{30}\;\pi5^2 $$