# Thread: problem in region bounded by a curve and a straight line

1. $\dfrac{\pi}{16} \left[49(7) - \frac{14(7)^2}{2} + \frac{7^3}{3}\right] - \left[49(3) - \frac{14(3)^2}{2} + \frac{3^3}{3} \right] - \pi\left[4(4) - \frac{4^2}{2}- 4(3) - \frac{3^2}{2} \right]$

This line should be:

$\dfrac{\pi}{16} \left[49(7) - \frac{14(7)^2}{2} + \frac{7^3}{3}\right] - \left[49(3) - \frac{14(3)^2}{2} + \frac{3^3}{3} \right] - \pi\left[(4(4) - \frac{4^2}{2})- (4(3) - \frac{3^2}{2}) \right]$

$\dfrac{\pi}{16} \left[49(7) - \frac{14(7)^2}{2} + \frac{7^3}{3}\right] - \left[49(3) - \frac{14(3)^2}{2} + \frac{3^3}{3} \right] - \pi\left[4(4) - \frac{4^2}{2}- 4(3) + \frac{3^2}{2} \right]$

Notice the plus sign.
Continue now.

2. i get:

$\dfrac{\pi}{16} \left[49(7) - \frac{14(7)^2}{2} + \frac{7^3}{3}\right] - \left[49(3) - \frac{14(3)^2}{2} + \frac{3^3}{3} \right] - \pi\left[4(4) - \frac{4^2}{2}- 4(3) + \frac{3^2}{2} \right]$

= $\frac{\pi}{16}\left[\frac{64}{3}\right] -\pi\left[\frac{1}{2}\right]$

= $\pi\frac{4}{3} - \pi\left[\frac{1}{2}\right]$

= $\frac{5}{6}\pi unit^3$

right?

3. $\pi\left[4(4) - \frac{4^2}{2}- 4(3) + \frac{3^2}{2} \right] = \pi (16 - 8 - 12 + 4.5) = 0.5\pi$

Giving final answer as $\frac56 \pi\ units^3$

4. okey sir..thanks very much..i really appreciate with you..thanks!!

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