Mathematics 9709 · Pure Mathematics 1

Area between curves & volumes of revolution

Two more common Paper 1 questions: the area between two graphs, and the volume from spinning a region around an axis.
Both use the same method. You just change what goes inside the integral.

Area between two graphs: integrate $(y_{\text{top}}-y_{\text{bot}})$ . Volume of revolution: $\pi\int y^2\,dx$ . So square $y$ first, then integrate and keep the $\pi$ .

Area between two graphs

For a region between an upper graph and a lower graph, integrate top minus bottom.
This one subtraction works for everything, even regions below the axis. That is because you measure the gap between the curves directly.

$A=\int_a^b\big(y_{\text{top}}-y_{\text{bot}}\big)\,dx$

The limits $a$ and $b$ are the $x$ -values where the graphs cross. So the steps are:

Find the crossings: set the two equations equal and solve. That gives your limits $a$ and $b$ (the calculator gives you the roots, just like in Quadratics).
Decide which graph is on top between the crossings — test one $x$ -value, or read it off a sketch.
Integrate top minus bottom between $a$ and $b$ .

Find where the graphs cross (a and b), then integrate the gap, top minus bottom, between them.

Worked example

Find the area between the line $y=x+2$ and the curve $y=x^2$

First find where they meet. Set them equal:

$x^2=x+2\;\Rightarrow\;x^2-x-2=0\;\Rightarrow\;x=-1,\ 2$

Which graph is on top? Test a point between the crossings, say $x=0$ . The line gives $2$ . The curve gives $0$ . So here the line is higher. Integrate $(\text{line})-(\text{curve})$ :

$A=\int_{-1}^{2}\big((x+2)-x^2\big)\,dx=\left[\tfrac{x^2}{2}+2x-\tfrac{x^3}{3}\right]_{-1}^{2}$

Answer

\tfrac{9}{2}\ \text{units}^2

A real Paper 1 question puts a root curve against a line, so the crossing equation is a quadratic in $\sqrt{x}$ : same steps.

Worked example

Find the exact area of the region bounded by the curve $y=5\sqrt{x}$ and the line $y=2x+2$ .

Find the crossings: $5\sqrt{x}=2x+2$ . This is a quadratic in $\sqrt{x}$ : squaring gives $4x^2-17x+4=0$ , so $x=\tfrac14$ or $x=4$ .

Between them the curve is on top (test $x=1$ : curve gives $5$ , line gives $4$ ). Integrate curve minus line:

$A=\int_{1/4}^{4}\big(5x^{\frac12}-2x-2\big)\,dx=\left[\tfrac{10}{3}x^{\frac32}-x^2-2x\right]_{1/4}^{4}$

Answer

\tfrac{45}{16}\ \text{units}^2

(9709/12 May/June 2022 Q6)

Curve and a straight line: subtract the trapezium

When the edges are a curve and a straight line, the textbook’s main method is often faster. Find the area under the curve. Then subtract the area under the line.
The line is straight, so the region under it is a trapezium (or triangle). You can find its area from the shape. You do not need a second integral.

$A=\int_a^b y_{\text{curve}}\,dx-\tfrac12(p+q)w$

Here $p$ and $q$ are the two vertical sides of the trapezium (the height of the line at $x=a$ and $x=b$ ). And $w=b-a$ is its width.
Mark schemes give full marks for this method.

Worked example

The curve $y=-x^2+8x-5$ and the line $y=x+1$ meet at $(1,\ 2)$ and $(6,\ 7)$ . Find the area between them.

The curve is above the line here. So take the area under the curve minus the trapezium under the line:

$A=\int_1^6\big(-x^2+8x-5\big)\,dx-\tfrac12(2+7)(5)$

The integral is the curve’s area. The trapezium has parallel sides $2$ and $7$ , and width $6-1=5$ :

$=\left[-\tfrac{x^3}{3}+4x^2-5x\right]_1^6-\tfrac{45}{2}=\tfrac{130}{3}-\tfrac{45}{2}$

Answer

\tfrac{125}{6}\ \text{units}^2

The top-minus-bottom integral $\int_1^6\big((-x^2+8x-5)-(x+1)\big)\,dx$ gives the same value, $\tfrac{125}{6}$ . So pick whichever is less work.
With a straight line, the trapezium method is usually easier. (9709-style)

Volumes of revolution

The chapter has four jobs: reverse differentiation, area, area between two curves, and now volume. They all use the same definite integral. A volume question is an area question where the height is $\pi y^2$ instead of $y$ . You change what goes inside the integral. The method stays the same.

Spin the region under a curve once around the $x$ -axis. It makes a solid shape.
Cut that solid into thin discs. Each one is a circle of radius $y$ , so its volume is $\pi y^2\,dx$ . Add them all up:

$V=\pi\int_a^b y^2\,dx$

Each slice is a disc of radius y, volume π y² dx. Spinning the whole region about the x-axis and adding the discs gives V = π ∫ y² dx.

Three steps — and the first one matters most:

Square $y$ first — write down $y^2$ before you do anything else. You cannot integrate $y$ and square afterwards.
Integrate $y^2$ between the limits, keeping the $\pi$ at the front the whole way.
Do top minus bottom and leave the answer as an exact multiple of $\pi$ .

Worked example

The region under $y=\sqrt{x}$ from $x=0$ to $4$ is rotated $360^\circ$ about the $x$ -axis. Find the volume.

Square $y$ first: $y^2=(\sqrt{x})^2=x$ . Then integrate:

$V=\pi\int_0^4 x\,dx=\pi\left[\tfrac{x^2}{2}\right]_0^4=\pi(8)$

Answer

8\pi

Rotating about the y-axis

The paper asks for rotation about the $y$ -axis just as often. It is the same method, but with $x$ and $y$ swapped.
Now the discs stack up sideways, each of radius $x$ . So you square $x$ and integrate with respect to $y$ :

$V=\pi\int_c^d x^2\,dy$

Two things change together:
Rearrange to get $x^2$ in terms of $y$ .
Use limits on $y$ , not on $x$ .

Worked example

The region bounded by $y=x^2$ , the $y$ -axis and the lines $y=2$ and $y=5$ is rotated $360^\circ$ about the $y$ -axis. Find the volume.

Rotating about the $y$ -axis, so use $V=\pi\int x^2\,dy$ . From $y=x^2$ the radius squared is just $x^2=y$ , and the limits are $y=2$ to $y=5$ :

$V=\pi\int_2^5 y\,dy=\pi\left[\tfrac{y^2}{2}\right]_2^5=\pi\left(\tfrac{25}{2}-\tfrac{4}{2}\right)$

Answer

\tfrac{21}{2}\pi

When the region is bounded by a curve AND a straight line

If the region is bounded by a curve and a straight line (not just the axis), rotate both and subtract. Take the solid from the curve minus the solid from the line.

When that line is horizontal or vertical, it makes a plain cylinder of volume $\pi r^2 h$ . So the answer is “outer solid minus cylinder”. This is exactly the M/J 2020 question in Now you try below. The rotated curve gives $12\pi$ . The line $x=1$ cuts out a cylinder of $4\pi$ . That leaves $8\pi$ .

Where the marks go

Common mistake

Integrating bottom minus top gives a negative area. Decide which graph is higher first. Do a quick sketch, or test one

x

-value between the crossings. Put that one on top. Do not assume the line is the higher one. In this paper the curve

y=-2x^2+8x+11

is above the line

y=8x+9

. (9709/12 Oct/Nov 2024 Q7)

Common mistake

For a volume you must square $y$ before you integrate. You cannot integrate

y

and square afterwards. And keep the

\pi

. The answers are exact multiples of

\pi

. (9709/12 May/June 2020 Q8)

Now you try

The points $(1,6)$ and $(3,2)$ lie on the curve $y=\dfrac{6}{x}$ . The region bounded by the curve, the line $y=2$ and the line $x=1$ is rotated 360° about the $y$ -axis. Find the volume. (9709/12 May/June 2020 Q8)

Your turn— tap to reveal the worked answer (9709/12 May/June 2020 Q8)

Rotating about the $y$ -axis, so use $V=\pi\int x^2\,dy$ with $x=\dfrac{6}{y}$ , between $y=2$ and $y=6$ :

$\pi\int_2^6\left(\tfrac{6}{y}\right)^2\,dy=\pi\big[-36y^{-1}\big]_2^6=12\pi$

The line $x=1$ cuts out a cylinder of radius $1$ and height $6-2=4$ . So subtract $\pi(1)^2(4)=4\pi$ :

Answer

V=12\pi-4\pi=8\pi

PreviousArea under a curve

Area between curves & volumes of revolution

Two more common Paper 1 questions: the area between two graphs, and the volume from spinning a region around an axis.
Both use the same method. You just change what goes inside the integral.

Area between two graphs: integrate $(y_{\text{top}}-y_{\text{bot}})$ . Volume of revolution: $\pi\int y^2\,dx$ . So square $y$ first, then integrate and keep the $\pi$ .

Area between two graphs

For a region between an upper graph and a lower graph, integrate top minus bottom.
This one subtraction works for everything, even regions below the axis. That is because you measure the gap between the curves directly.

$A=\int_a^b\big(y_{\text{top}}-y_{\text{bot}}\big)\,dx$

The limits $a$ and $b$ are the $x$ -values where the graphs cross. So the steps are:

Find the crossings: set the two equations equal and solve. That gives your limits $a$ and $b$ (the calculator gives you the roots, just like in Quadratics).
Decide which graph is on top between the crossings — test one $x$ -value, or read it off a sketch.
Integrate top minus bottom between $a$ and $b$ .

Find where the graphs cross (a and b), then integrate the gap, top minus bottom, between them.

Worked example

Find the area between the line $y=x+2$ and the curve $y=x^2$

First find where they meet. Set them equal:

$x^2=x+2\;\Rightarrow\;x^2-x-2=0\;\Rightarrow\;x=-1,\ 2$

Which graph is on top? Test a point between the crossings, say $x=0$ . The line gives $2$ . The curve gives $0$ . So here the line is higher. Integrate $(\text{line})-(\text{curve})$ :

$A=\int_{-1}^{2}\big((x+2)-x^2\big)\,dx=\left[\tfrac{x^2}{2}+2x-\tfrac{x^3}{3}\right]_{-1}^{2}$

Answer

\tfrac{9}{2}\ \text{units}^2

A real Paper 1 question puts a root curve against a line, so the crossing equation is a quadratic in $\sqrt{x}$ : same steps.

Worked example

Find the exact area of the region bounded by the curve $y=5\sqrt{x}$ and the line $y=2x+2$ .

Find the crossings: $5\sqrt{x}=2x+2$ . This is a quadratic in $\sqrt{x}$ : squaring gives $4x^2-17x+4=0$ , so $x=\tfrac14$ or $x=4$ .

Between them the curve is on top (test $x=1$ : curve gives $5$ , line gives $4$ ). Integrate curve minus line:

$A=\int_{1/4}^{4}\big(5x^{\frac12}-2x-2\big)\,dx=\left[\tfrac{10}{3}x^{\frac32}-x^2-2x\right]_{1/4}^{4}$

Answer

\tfrac{45}{16}\ \text{units}^2

(9709/12 May/June 2022 Q6)

Curve and a straight line: subtract the trapezium

When the edges are a curve and a straight line, the textbook’s main method is often faster. Find the area under the curve. Then subtract the area under the line.
The line is straight, so the region under it is a trapezium (or triangle). You can find its area from the shape. You do not need a second integral.

$A=\int_a^b y_{\text{curve}}\,dx-\tfrac12(p+q)w$

Here $p$ and $q$ are the two vertical sides of the trapezium (the height of the line at $x=a$ and $x=b$ ). And $w=b-a$ is its width.
Mark schemes give full marks for this method.

Worked example

The curve $y=-x^2+8x-5$ and the line $y=x+1$ meet at $(1,\ 2)$ and $(6,\ 7)$ . Find the area between them.

The curve is above the line here. So take the area under the curve minus the trapezium under the line:

$A=\int_1^6\big(-x^2+8x-5\big)\,dx-\tfrac12(2+7)(5)$

The integral is the curve’s area. The trapezium has parallel sides $2$ and $7$ , and width $6-1=5$ :

$=\left[-\tfrac{x^3}{3}+4x^2-5x\right]_1^6-\tfrac{45}{2}=\tfrac{130}{3}-\tfrac{45}{2}$

Answer

\tfrac{125}{6}\ \text{units}^2

The top-minus-bottom integral $\int_1^6\big((-x^2+8x-5)-(x+1)\big)\,dx$ gives the same value, $\tfrac{125}{6}$ . So pick whichever is less work.
With a straight line, the trapezium method is usually easier. (9709-style)

Volumes of revolution

Spin the region under a curve once around the $x$ -axis. It makes a solid shape.
Cut that solid into thin discs. Each one is a circle of radius $y$ , so its volume is $\pi y^2\,dx$ . Add them all up:

$V=\pi\int_a^b y^2\,dx$

Each slice is a disc of radius y, volume π y² dx. Spinning the whole region about the x-axis and adding the discs gives V = π ∫ y² dx.

Three steps — and the first one matters most:

Square $y$ first — write down $y^2$ before you do anything else. You cannot integrate $y$ and square afterwards.
Integrate $y^2$ between the limits, keeping the $\pi$ at the front the whole way.
Do top minus bottom and leave the answer as an exact multiple of $\pi$ .

Worked example

The region under $y=\sqrt{x}$ from $x=0$ to $4$ is rotated $360^\circ$ about the $x$ -axis. Find the volume.

Square $y$ first: $y^2=(\sqrt{x})^2=x$ . Then integrate:

$V=\pi\int_0^4 x\,dx=\pi\left[\tfrac{x^2}{2}\right]_0^4=\pi(8)$

Answer

8\pi

Rotating about the y-axis

The paper asks for rotation about the $y$ -axis just as often. It is the same method, but with $x$ and $y$ swapped.
Now the discs stack up sideways, each of radius $x$ . So you square $x$ and integrate with respect to $y$ :

$V=\pi\int_c^d x^2\,dy$

Two things change together:
Rearrange to get $x^2$ in terms of $y$ .
Use limits on $y$ , not on $x$ .

Worked example

The region bounded by $y=x^2$ , the $y$ -axis and the lines $y=2$ and $y=5$ is rotated $360^\circ$ about the $y$ -axis. Find the volume.

Rotating about the $y$ -axis, so use $V=\pi\int x^2\,dy$ . From $y=x^2$ the radius squared is just $x^2=y$ , and the limits are $y=2$ to $y=5$ :

$V=\pi\int_2^5 y\,dy=\pi\left[\tfrac{y^2}{2}\right]_2^5=\pi\left(\tfrac{25}{2}-\tfrac{4}{2}\right)$

Answer

\tfrac{21}{2}\pi

When the region is bounded by a curve AND a straight line

If the region is bounded by a curve and a straight line (not just the axis), rotate both and subtract. Take the solid from the curve minus the solid from the line.

Where the marks go

Common mistake

Integrating bottom minus top gives a negative area. Decide which graph is higher first. Do a quick sketch, or test one

x

-value between the crossings. Put that one on top. Do not assume the line is the higher one. In this paper the curve

y=-2x^2+8x+11

is above the line

y=8x+9

. (9709/12 Oct/Nov 2024 Q7)

Common mistake

For a volume you must square $y$ before you integrate. You cannot integrate

y

and square afterwards. And keep the

\pi

. The answers are exact multiples of

\pi

. (9709/12 May/June 2020 Q8)

Now you try

Your turn— tap to reveal the worked answer (9709/12 May/June 2020 Q8)

Rotating about the $y$ -axis, so use $V=\pi\int x^2\,dy$ with $x=\dfrac{6}{y}$ , between $y=2$ and $y=6$ :

$\pi\int_2^6\left(\tfrac{6}{y}\right)^2\,dy=\pi\big[-36y^{-1}\big]_2^6=12\pi$

The line $x=1$ cuts out a cylinder of radius $1$ and height $6-2=4$ . So subtract $\pi(1)^2(4)=4\pi$ :

Answer

V=12\pi-4\pi=8\pi