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.

Re: Need Solution Today..

Quote:

Originally Posted by

**faizanmax** **Question#1**

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Evaluate the integral.

Attachment 28738
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Split the range of integration at the "corner"$\displaystyle t=5/3$, where $\displaystyle 3t-5$ changes sign, then the integral becomes:

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

I leave the rest to you.

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Re: Need Solution Today..

Quote:

Originally Posted by

**faizanmax** __________________________________________________ ___________________________

**Question#2**

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Find the volume of solid obtained by revolving around the y-axis the plane area between the graph

Attachment 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 $\displaystyle x$-axis of that part of the curve above the $\displaystyle x$-axis. This is the part of the curve for which $\displaystyle 0\le x \le 1$.

Now your volume of rotation is:

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

and the rest you can do for yourself.

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