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Thread: integration for area&volume

  1. #1
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    integration for area&volume

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

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

    for qn 1 :
    i can do the first part....my ans is $\displaystyle ln2$
    i can do the second part too by using integration by parts....my ans is $\displaystyle \frac{\pi \sqrt3}{3}-ln2$
    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
    $\displaystyle \int^2 _0 \frac{5}{(x+1)(x^2+4)} dx$
    $\displaystyle =5 \int^2 _0 \frac{1}{x^3 + x^2+ 4x+4}$

    then ...i'm stuck....
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  2. #2
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    Dec 2010
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    Hi there,

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

    Expand and gather all like terms:
    $\displaystyle
    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):
    $\displaystyle \int\frac{1}{x+1}+\int\frac{-x+1}{x^2+4}$

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

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

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

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

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


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

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

    This is exact. You could simplify the natural log further if you wanted to.
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  3. #3
    Senior Member
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    Nov 2010
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    Clarksville, ARk
    Posts
    398

    Re: integration for area&volume

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


    for qn 1 :
    i can do the first part....my ans is $\displaystyle \ln2$
    i can do the second part too by using integration by parts....my ans is $\displaystyle \frac{\pi \sqrt3}{3}-\ln2$
    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 $\displaystyle \int_0^{\sqrt3} tan^{-1} (y)\ dy$ 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 $\displaystyle \displaystyle \text{v}= \left(\pi y_2^{\,2}-\pi y_1^{\,2}\right)\Delta x\,.$ In the case of region S, $\displaystyle \displaystyle y_1=\tan(x)$ and $\displaystyle \displaystyle y_2=\sqrt{3}.$ Letting Δx be very small, it becomes dx. Integrate this over x to get the volume.

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

    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 $\displaystyle \displaystyle \text{v}= 2\pi y\left( x_2- x_1\right)\Delta y\,.$ In the case of region S, $\displaystyle \displaystyle x_1=0$ and $\displaystyle \displaystyle x_2=\tan^{-1}(y).$ Letting Δy be very small, it becomes dy. Integrate this over y to get the volume.

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