Find volume of the solid revolved around $y=4$

Question

Consider the function $f(x)=\sqrt{1-x}$

The region is bounded by $y=0$ and $x=0$

So I write the integral as so $$ \pi \int_0^44^2-(\sqrt{1-x})^2dx$$ Solving it gives the answer $68\pi$

I don't have the answer given in the back of my textbook and I was wondering if I setup the integral correctly.

Answer 1

The integral with the washer method should be

$$ \pi \int_0^1 [4^2-(4-\sqrt{1-x})^2]dx$$

where 4 and $4-\sqrt{1-x}$ are the outer and inner radius of the ring, respectively. Note that the integral is over $x$, which has its upper bound $x=1$ given by $0=\sqrt{1-x}$.

Answer 2

Using cylindrical shells you end up with an easier integral. The formula is

$$ \int_0^1 2 \pi ( 4 - y) (1 - y^2) dy = \frac{29 \pi}{6} $$

Note the following. The cylindrical shell is seated on the y axis. Therefore the radius $r$, height $h$ and thickness of the cylindrical shell respectively are:

$$ r = (4 - y), \quad h = (1 - x) \quad \text{and}\; \mathrm{thickness} = dy $$

Since the eventual integral is going to be in terms of $dy$ we need to solve for $x$ in the equation for $h$. We get $x = 1 - y^2$. Once again our integral is in terms of $y$. So our bounds on the integral too needs to be in terms of $y$. The bounds on $y$ are $0$ and $1$.

Note that this gives the same answer as that of @Quanto. His/her solution requires that you expand the term $(4 - \sqrt{1 -x})^2$. Which results in quite a nasty integral that still has a square root term in it.

Drawing a picture and trying to figure out the various radii, heights and thicknesses (if you are using cylindrical shells) helps.