Find the volume of the solid of revolution obtained by rotating the region bounded by the curves $x=0$, $y=0$ and $x+y=1$ around the line $x+y=2$.
Answer:
We have to rotate the red triangle about the green line and hence we have to find the volume generated.
Area of the triangle is $A=\frac{1}{2}$.
Let $P=(x,y)$ be a point on the lien $x+y=2$.
Inner radius=Distance between the line $x+y=1$ and $x+y=2$ is $R=\frac{x+y-2}{\sqrt{x^2+y^2}}$,
Outer radius=distance between $P=(x,y)$ to $(0,0)$ is $r=\sqrt{x^2+y^2}$.
The volume is $=A \int_{1}^{2} [r^2-R^2]dx$
So far am I right?
Best Answer
I think you completely misunderstood the problem. If you would rotate the region bounded by $x=0$, $y=0$, $x+y=2$ around the $x+y=2$ line you would get two cones that share the base, with $x+y=2$ being the axis of rotation. Now if you take only the region up to $x+y=1$, you get a hole along the axis of rotation in the above object. For simplicity, let's choose the new origin $O'$ to be the point $(1,1)$, where the perpendicular from $(0,0)$ intersects $x+y=2$. You can compute the volume by either cylindrical shell method or the washer method.
Using the washer method: you have washers that have the outer point either on the $x$ or $y$ axis, and the inner point on the $x+y=1$ line. The height of the washer is $dz$, where $z$ varies between $-\frac{\sqrt{2}}2$ and $ \frac{\sqrt{2}}2$. The $z$ axis is along $x+y=2$. The inner radius is always $\frac{\sqrt 2}2$. Due to the symmetry of the figure, the volume is twice that of the figure, you can calculate the integral only from $z=0$, then multiply it by $2$. The outer radius at $z=0$ is $\sqrt 2$ and goes down to $ \frac{\sqrt{2}}2$ when $z= \frac{\sqrt{2}}2$. This means that the inner radius is $\sqrt 2-z$. So the volume is $$V=2\int_0^{\frac{\sqrt 2}2}\pi\left[(\sqrt 2-z)^2-\left(\frac{\sqrt 2}2\right)^2\right]dz$$