1. ## integration for area&volume

1. O is the origin and A is the point on the curve y=tan x where $x=\frac{\pi}{3}.$
Calculate the area of the region R enclosed by the arc OA, the x-axis and the line $x=\frac{\pi}{3}$, giving your answer in EXACT form. Hence, find $\int^{\sqrt3} _0 tan^{-1} y dy$ in EXACT form.
Then, the region S is enclosed by the arc OA, the y-axis ad the line $y=\sqrt3$. Find the volume of the solid of revolution formed when S is rotated through $2\pi$ about the x-axis, giving your answer in EXACT form.

2. Give your answer in EXACT form for $\int^2 _0 \frac{5}{(x+1)(x^2+4)} dx$

for qn 1 :
i can do the first part....my ans is $ln2$
i can do the second part too by using integration by parts....my ans is $\frac{\pi \sqrt3}{3}-ln2$
then I don't know how to do the last part. if it rotated about y-axis , then i know, but it rotated about the x-axis.

for qn 2:
what i tried to do was
$\int^2 _0 \frac{5}{(x+1)(x^2+4)} dx$
$=5 \int^2 _0 \frac{1}{x^3 + x^2+ 4x+4}$

then ...i'm stuck....

2. Hi there,

For question #2, we use integration by partial fractions.
$
\frac{5}{(x+1)(x^2+4)} = \frac{A}{x+1}+\frac{Bx+C}{x^2+4}
$

By multiplying each side by the denominator on the left, we get:
$
5=A(x^2+4)+(Bx+C)(x+1)
$

Expand and gather all like terms:
$
5=Ax^2+Bx^2+Bx+Cx+4A+C
$

Equate coefficients on the right to those on the left (x^2 and x don't exist, so their coefficients must be 0):
for x^2, (A+B)=0
for x, (B+C)=0
for constant, (4A+C)=5
By solving this system, we find A=1, B=-1, C=1.

Now we can integrate (from the very first equation, plugging in A,B,C):
$\int\frac{1}{x+1}+\int\frac{-x+1}{x^2+4}$

The first integral is easy, $\ln{(x+1)}$

The second requires us to split up the integral:
$\int\frac{-x+1}{x^2+4}=\int\frac{1}{x^2+4}-\int\frac{x}{x^2+4}$

This time, the first integral is harder, we notice that this fits the formula for the derivative of arctan, with a=2:
$\int\frac{1}{(x^2+a^2)}=\frac{1}{a}arctan(\frac{x} {a})$

so, the first part of this second integral is equal to:
$\frac{1}{2}arctan(\frac{x}{2})$

And the second part is easy, $-\frac{\ln{(x^2+4)}}{2}$

This leaves us with a final indefinite integral of:
$
ln{(x+1)}+\frac{1}{2}arctan(\frac{x}{2})-\frac{\ln{(x^2+4)}}{2}
$

Now evaluate this from 0 to 2.
$\ln{3}+\frac{\Pi}{8}-\frac{\ln{8}}{2}+\frac{\ln{4}}{2}$

This is exact. You could simplify the natural log further if you wanted to.

3. ## Re: integration for area&volume

Originally Posted by wintersoltice
1. O is the origin and A is the point on the curve y=tan x where $x=\frac{\pi}{3}.$
Calculate the area of the region R enclosed by the arc OA, the x-axis and the line $x=\frac{\pi}{3}$, giving your answer in EXACT form. Hence, find $\int^{\sqrt3} _0 tan^{-1} y dy$ in EXACT form.
Then, the region S is enclosed by the arc OA, the y-axis ad the line $y=\sqrt3$. Find the volume of the solid of revolution formed when S is rotated through $2\pi$ about the x-axis, giving your answer in EXACT form.

for qn 1 :
i can do the first part....my ans is $\ln2$
i can do the second part too by using integration by parts....my ans is $\frac{\pi \sqrt3}{3}-\ln2$
then I don't know how to do the last part. if it is rotated about y-axis , then i know, but it is rotated about the x-axis.

then ...i'm stuck....

For question 1:

The integral $\int_0^{\sqrt3} tan^{-1} (y)\ dy$ gives the area of region S, not region R.

Your answers for the area of region R and the area of region S are correct.

The methods for revolving about the x-axis are essentially the same as for revolving about the y-axis. You can use disks/washers or cylindrical shells.

Disks/Washers: In this case washers. Form a washer from a rectangle which extends from y_1 to y_2 vertically and has a of width, Δx. Rotate the rectangle about the x-axis. The volume of the washer is $\displaystyle \text{v}= \left(\pi y_2^{\,2}-\pi y_1^{\,2}\right)\Delta x\,.$ In the case of region S, $\displaystyle y_1=\tan(x)$ and $\displaystyle y_2=\sqrt{3}.$ Letting Δx be very small, it becomes dx. Integrate this over x to get the volume.

$\displaystyle V=\pi\int_0^{\pi/3} \left(\sqrt{3}\right)^2-\tan^2(x)\,dx$

Cylindrical shells:Form a cylindrical shell from a rectangle which extends from x_1 to x_2 horizontally and has a of height, Δy, located a distance of y above the x-axis.. Rotating this about the x-axis gives a cylindrical shell with a volume of $\displaystyle \text{v}= 2\pi y\left( x_2- x_1\right)\Delta y\,.$ In the case of region S, $\displaystyle x_1=0$ and $\displaystyle x_2=\tan^{-1}(y).$ Letting Δy be very small, it becomes dy. Integrate this over y to get the volume.

$\displaystyle V=2\pi\int_0^{\sqrt{3}} y\tan^{-1}(y)\,dy$