# Finding Volume using Integration for Area Bound by Two Curves

I am trying to find the volume of the solid formed by rotating the region of the region bounded by $$y=x$$ and $$y=\sqrt x$$

about the line $$x=2$$

I don't understand how to set up this integral because it is not rotated around the y-axis. I also don't understand how to find the area for the integral.

Thank you!

• I suppose it is for $0\le x\le 1$? Commented Mar 12, 2016 at 0:30
• That's what I don't understand. Should I only be calculating the area of that fragment or should I be considering the area from [0,2] ?
– Ana
Commented Mar 12, 2016 at 0:31
• Anyway, in the second case, you have to split into two parts: from $0$ to $1$ and from $1$ to $2$. Commented Mar 12, 2016 at 0:33
• I still don't understand how to set up the integrals. I know I should be setting the integral with respect to y, but I don't know how to set up the area of each cross section or how the fact that I'm rotating about x=2 rather than the y axis plays in.
– Ana
Commented Mar 12, 2016 at 0:35
• The limits of integration will be from 0 to 1, because that is where the region exists. But it can revolve about an axis that is outside that region. Use cylindrical shells the radius = (2-x). Commented Mar 12, 2016 at 0:40

First sketch the curves $y=x$ and $y=\sqrt{x}$. Find the intersection points of the two curves by solving simultaneously $y=x$ and $y=\sqrt{x}$, that is $x=\sqrt{x}$, equivalently $x-\sqrt{x}=0$, hence $\sqrt{x}(\sqrt{x}-1)=0$, with solutions $x=0$ and $x=1$. Since $y=x$, the coordinates of the intersection points are $(0,0)$ and $(1,1)$. (Alternatively, $x=\sqrt{x}$, hence $x^2=x$, this $x^2-x=0$ same as $x(x-1)=0$ giving solutions $x=0$ and $x=1$.)

You need to revolve it about the line $x=2$ (which is a vertical line). Here is the result.

You may take integral either with respect to $x$, $\int..dx$, or with respect to $y$, $\int..dy$.

Note when you project your region onto the $x$-axis you get the interval $x\in[0,1]$. It so happens for this particular problem that if you project onto the $y$-axis you get the interval $y\in[0,1]$. Even if the interval is $[0,1]$ in both cases, it helps to think of one of them as lying on the $x$-axis (from $x=0$ to $x=1$) while the other is lying on the $y$-axis (from $y=0$ to $y=1$).

Say, first try $\int..dx$. Since $x\in[0,1]$, this will be $\int_0^1..dx$. Take a generic $x\in[0,1]$ and draw the line that goes with it, this will be a vertical line, and it cuts the given region in a certain vertical line segment. Revolve this line segment about the given axis of revolution, $x=2$. The result is a cylinder, or a cylindrical shell (just the side, not including top or bottom).

You need to find the area of this cylinder (as depending on $x$) and put it inside your integral $\int_0^1..dx$.
The area of a cylinder is $A=circumference\cdot height$.
The circumference is $circumference=2\pi\cdot radius$.
Hence the area is $A=2\pi\cdot radius\cdot height$.
Take a look at the picture below. The radius is the distance between the axis of rotation $x=2$, and the particular $x\in[0,1]$ that determines this cylindrical shell. You could think of the latter $x$ as the variable of integration, the $x$ within the $dx$. It stands on its own, and the area will be expressed in terms of this $x$. I will make it large $\Large x$ to make it stand out. So the radius is the distance from the line $x=2$ to this $\Large x$, and since ${\Large x}<2$, this distance is $2-{\Large x}$.

So far we get $radius = 2- {\Large x}$. We also need to find the height of the shell and express it in terms of this ${\Large x}$. The height, on the next picture, is a vertical line segment, so it will be the difference between the $y$-coordinates of the top point of the segment and the bottom point of segment. The top point lies on the curve $y=\sqrt x$ and the bottom point on the curve $y=x$. So these two points have coordinates $({\Large x},y_{top}=\sqrt{\Large x})$ and $({\Large x},y_{bottom}={\Large x})$. The height of the shell equals
$height = y_{top} - y_{bottom} = \sqrt{\Large x} - {\Large x}$.

Therefore the area of the shell is
$A=2\pi\cdot radius\cdot height = 2\pi\cdot (2-{\Large x})\cdot (\sqrt{\Large x} - {\Large x})$.

The volume is $V=\int area = \int_0^1 A dx$. So we obtain

$V=\displaystyle\int_0^1 2\pi(2-{\Large x})\cdot (\sqrt{\Large x} - {\Large x}) d{\Large x}$.
It remains to compute the above integral:
$\displaystyle \int_0^1 2\pi(2-x)\cdot (\sqrt{x} - x) dx = \displaystyle 2\pi\int_0^1 (2x^{\frac12}-2x-x^{\frac32}+x^2) dx =$
$\displaystyle 2\pi (\frac{2x^{\frac32}}{\frac32}- \frac{2x^2}2-\frac{x^{\frac52}}{\frac52}+\frac{x^3}3) \Bigr|_0^1 = 2\pi(\frac43-1-\frac25+\frac13)=$
$\displaystyle 2\pi(\frac{20-15-6+5}{15})= 2\pi(\frac{4}{15})= \frac{8\pi}{15}$.

Say, now try $\int..dy$. Since $y\in[0,1]$ (on the vertical axis), this will be $\int_0^1..dy$. Take a generic $y\in[0,1]$ and draw the line that goes with it, this will be a horizontal line, and it cuts the given region in a certain horizontal line segment. Revolve this line segment about the given axis of revolution, $x=2$. The result is a washer, also called annulus, a larger disk from which a smaller disk (with the same center) is removed. See the picture below.

The area of a washer is the area of the larger disk minus the area of the smaller disk. The area of a disk (that is, the area bounded by a circle) is $area\ of \ circle= \pi r^2$, where $r$ denotes its radius. Here we have a smaller radius (of the smaller disk) and a larger radius (or the larger disk).

The horizontal line through the ${\Large y}$ (the variable of integration) cuts the given region at a horizontal line segment with endpoints $(x_{outer},{\Large y})$ and $(x_{inner},{\Large y})$ as shown in the picture.

The small radius is the distance from the point $(x_{inner},{\Large y})$ to the axis of revolution $x=2$. This distance is represented by the horizontal red line segment on the picture below, and is the difference between the $x$-coordinates of its endpoints. The larger $x$-coordinate is $x=2$, and the smaller $x$-coordinate is $x_{inner}$ (that goes with the inner disk). Therefore the small radius is
$r_{small}=2-x_{inner}$. Note that $(x_{inner},{\Large y})$ belongs to the line $y=x$, hence ${\Large y}=x_{inner}$, which solved for $x_{inner}$ gives $x_{inner}={\Large y}$. Thus $r_{small}=2-x_{inner}=2- {\Large y}$.

Similarly, the large radius is the distance from the point $(x_{outer},{\Large y})$ to the axis of revolution $x=2$. This distance is the difference between the following $x$-coordinates: $x=2$ (the axis of revolution) and $x_{outer}$ (that goes with the outer disk). Therefore the large radius is $r_{large}=2-x_{outer}$. Note that $(x_{outer},{\Large y})$ belongs to the line $y=\sqrt x$, hence ${\Large y}=\sqrt{x_{outer}}$, which solved for $x_{outer}$ gives $x_{outer}={\Large y}^2$. Thus $r_{large}=2-x_{outer}=2- {\Large y}^2$.

Thus the area of the larger disk is $A_{large}=\pi (r_{large})^2 = \pi (2- {\Large y}^2)^2$.
The area of the smaller disk is $A_{small}=\pi (r_{small})^2 = \pi (2- {\Large y})^2$.
The area of the washer is the difference:
$A_{washer}=A_{large} - A_{small} = \pi (2- {\Large y}^2)^2 - \pi (2- {\Large y})^2$.
Take this area and plug it into the formula $Volume =\int Area$. More precisely, $V=\displaystyle \int_0^1 A_{washer}\ d{\Large y} = \int_0^1 \pi (2- {\Large y}^2)^2 - \pi (2- {\Large y})^2\ d{\Large y} =$ $\displaystyle \pi \int_0^1 (2-y^2)^2 - (2-y)^2\ dy = \pi \int_0^1 4- 4y^2 + y^4 - (4- 4y+y^2)\ dy =$ $\displaystyle \pi \int_0^1 4- 4y^2 + y^4 - 4+ 4y-y^2\ dy = \pi \int_0^1 y^4 - 5y^2 + 4y\ dy =$ $\displaystyle \pi ( \frac{y^5}5 - \frac{5y^3}3 + \frac{4y^2}2 ) \Bigr|_0^1 = \pi ( \frac{1}5 - \frac{5}3 + \frac{4}2 ) = \pi ( \frac{6 - 50 + 60}{30} ) = \pi ( \frac{16}{30} ) = \frac{8\pi}{15}$.

In this example the axis of revolution was vertical. In this case $\int..dx$ goes with cylindrical shells, while $\int..dy$ goes with washers.

On the other hand, if the axis of revolution were horizontal, of the from $y=b$, then the same methods as above would show that in this case $\int..dx$ would go with washers, while $\int..dy$ would go with cylindrical shells. The main ideas remain the same:

$V = \int A$. That is, volume is integral of area.

Area of cylinder is $2\pi\cdot radius \cdot height$.

Area of washer is $\pi(r_{large})^2-\pi(r_{small})^2$.

To determine the height, and the radii, for horizontal segments subtract the smaller $x$-coordinate from the larger. For vertical segments subtract the smaller $y$-coordinate from the larger.

If you take $\int..dx$ there there is a generic $\Large x$ and you need to express the area in terms of this $\Large x$. Lengths of horizontal segments might be easier to determine, while for lengths of vertical segments you might need to express $y_{top}$ and $y_{bottom}$ in terms of this $\Large x$.

If you take $\int..dy$ there there is a generic $\Large y$ and you need to express the area in terms of this $\Large y$. This time lengths of vertical segments might be easier to determine, while for lengths of horizontal segments you might need to express $x_{left}$ and $x_{right}$ in terms of this $\Large y$.

If the limits of integration are not given, then you would need to start by determining the region that is bounded by the given curves (by solving some equations simultaneously). Sometimes, depending on the region, $\int..dx$ might be a better choice, sometimes $\int..dy$ might be better, and sometimes both $\int..dx$ and $\int..dy$ might work equally well. (Sometimes you may need to split the integral into two integrals, and this may be avoided for a different choice of the variable of integration. For examle, for the region pictured below, $\int..dy$ would likely work better than $\int..dx$.) Of course, once you set up the integral correctly, you also need to evaluate it, and to make sure you distribute negative sign appropriately, and make computations carefully. It is easy, you just might need more practice.

Begin by creating a sketch of the graph. You need to get over the fact that we are rotating about a vertical axis. We see that in the interval $(0, 1),$ $\sqrt{x} > x.$ (Notice we do NOT consider the interval $(1, 2)$ because the problem said the region bounded by the two functions, not $x = 2.$)

Use the method of cylindrical shells, because the functions are already given in terms of $x.$ Our bounds are $x = 0$ and $x = 1.$ At any point $x$ between these bounds, the radius of the cylinder is is $2 - x,$ and the height of the cylinder is $\sqrt{x} - x.$ Our integral is $$V = 2\pi \int_{0}^{1} (2 - x)\left(\sqrt{x} - x\right) dx$$ $$= 2\pi \int_{0}^{1} \left(x^{2} - x^{\frac{3}{2}} - 2x + 2x^{\frac{1}{2}}\right) dx$$ $$= 2\pi \left[\frac{1}{3}x^{3} - \frac{2}{5}x^{\frac{5}{2}} - x^{2} + \frac{4}{3}x^{\frac{3}{2}}\right]_{0}^{1}$$ $$= 2\pi \left[\frac{1}{3} - \frac{2}{5} - 1 + \frac{4}{3}\right]$$ $$= 2\pi \left(\frac{4}{15}\right)$$ $$= \boxed{\frac{8\pi}{15}}.$$