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Math Help - Need Solution Today..

  1. #1
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    Need Solution Today..

    Question#1
    __________________________________________________ ___________________________
    Evaluate the integral.

    Need Solution Today..-mth.png

    Question#2
    __________________________________________________ ___________________________
    Find the volume of solid obtained by revolving around the y-axis the plane area between the graph Need Solution Today..-mth2.pngand the x-axis.

    Please give the solution with all possible necessary steps..
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  2. #2
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    Re: Need Solution Today..

    Sorry but that is just not going to happen. We are not a homework service. If you are asking for help you are expected to show some effort, which means that you have attempted the problems, shown us everything you have tried and exactly where you are stuck. Then we can give you guidance.
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  3. #3
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    Re: Need Solution Today..

    Quote Originally Posted by faizanmax View Post
    Question#1
    __________________________________________________ ___________________________
    Evaluate the integral.

    Click image for larger version. 

Name:	mth.PNG 
Views:	19 
Size:	813 Bytes 
ID:	28738


    __________________________________________________ ___________________________

    Split the range of integration at the "corner"  t=5/3, where 3t-5 changes sign, then the integral becomes:

    \int_0^3 |3t-5|\; dt = \int_0^{5/3} (5-3t) \; dt +  \int_{5/3}^3 (3t-5) \; dt

    I leave the rest to you.

    .
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  4. #4
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    Re: Need Solution Today..

    Quote Originally Posted by faizanmax View Post
    __________________________________________________ ___________________________


    Question#2
    __________________________________________________ ___________________________
    Find the volume of solid obtained by revolving around the y-axis the plane area between the graph Click image for larger version. 

Name:	mth2.PNG 
Views:	20 
Size:	431 Bytes 
ID:	28740and the x-axis.

    Please give the solution with all possible necessary steps..
    This is ambiguous, draw a graph of the function and you will see why. I will assume it means find the volume of rotation anout the x-axis of that part of the curve above the x-axis. This is the part of the curve for which 0\le x \le 1.

    Now your volume of rotation is:

     V=\int_{x=0}^1 \pi y^2 \; dx

    and the rest you can do for yourself.

    .
    Last edited by zzephod; July 5th 2013 at 10:59 PM.
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