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Integral Calculus and its applications
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The area of the region bounded by the curve $xy=1, x$-axis$\;x=1.$ Find the volume of the solid generated by revolving the area mentioned about $x$-axis.
tnstate
class12
bookproblem
ch7
sec-1
exercise7-4
p118
q16
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asked
Apr 30, 2013
by
poojasapani_1
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1 Answer
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Toolbox:
Area bounded by the curve $t=f(x),$ the x-axis and the ordinates $x=a,x=b$ is $\int \limits_a^b f(x) dx $ or $ \int \limits _a^b y dx $
If the curve lies below the x-axis for $a \leq x \leq b,$ then the area is $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
Step 1:
$\pi \large\frac{r^2}{h^2}\bigg[\frac{x^3}{3}\bigg]_0^h$
$\quad=\large\frac{\pi}{3}\frac{r^2}{h^2}.$$h^3$
$\quad=\large\frac{1}{3} $$\pi r^2 h \;units$
Step 2:
The volume of the solid generated by the shaded region about the x axis is $v=\int \limits_0^1 \pi y^2 dx$
Step 3:
http://clay6.com/mpaimg/tn7.416.JPG
$\qquad=\pi \int \limits_0^1 \large\frac{1}{x^2}\;$$dx$
$\qquad=\pi \bigg[\large\frac{-1}{x}\bigg]_0^1$
answered
Aug 16, 2013
by
meena.p
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