# `y = (1/4)x^2, y = 5 - x^2` Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. (about the x-axis)

`y=x^2/4,y=5-x^2`

Refer the attached image. Graph of `y=x^2/4` is plotted in red color and graph of `y=5-x^2` is plotted in blue color.

From the graph, the curves intersects at x=-2 , x=2.

Using washer method,

Cross sectional area of washer A(x)=`pi((5-x^2)^2-(x^2/4)^2)`

`A(x)=pi(25+x^4-10x^2-x^4/16)`

`A(x)=pi(15/16x^4-10x^2+25)`

Volume of...

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`y=x^2/4,y=5-x^2`

Refer the attached image. Graph of `y=x^2/4` is plotted in red color and graph of `y=5-x^2` is plotted in blue color.

From the graph, the curves intersects at x=-2 , x=2.

Using washer method,

Cross sectional area of washer A(x)=`pi((5-x^2)^2-(x^2/4)^2)`

`A(x)=pi(25+x^4-10x^2-x^4/16)`

`A(x)=pi(15/16x^4-10x^2+25)`

Volume of the solid obtained by rotating the region bounded by the curves about the x=axis ,V=`int_(-2)^2A(x)dx`

`V=int_(-2)^2pi(15/16x^4-10x^2+25)dx`

`V=pi[15/16*x^5/5-10*x^3/3+25x]_(-2)^2`

`V=pi[3/16*x^5-10/3*x^3+25x]_(-2)^2`

`V=pi((3/16*2^5-10/3*2^3+25*2)-(3/16*(-2)^5-10/3*(-2)^3+25*(-2)))`

`V=pi((6-80/3+50)-(-6+80/3-50))`

`V=pi(6-80/3+50+6-80/3+50)`

`V=pi(112-160/3)`

`V=pi(336-160)/3`

`V=pi(176/3)`

`V=(176pi)/3`

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