1)find b such that the line y=b divides the region bounded by the graphs pf the
ID: 2871518 • Letter: 1
Question
1)find b such that the line y=b divides the region bounded by the graphs pf the two equations into two regions of equal area. round answers to 3 decimal places. y=49-x^2, y=0....b=?
2)consider the following f(x) = 1/7 (1+x^2), g (x)= 1/14 x^2... a) find the area of the region analytically (round answers to 3 decimal places) b) use the integration capabilities of the graphing utility to verify ur results. round answers to 3 decimal places.
3)Use the shell method to setup and evaluate the integral that gives the volume of the solid that generated by revolving the plane region about the x axis. y=4-x
Explanation / Answer
1)find b such that the line y=b divides the region bounded by the graphs pf the two equations into two regions of equal area. round answers to 3 decimal places. y=49-x^2, y=0....b=
y=49-x^2, y=0
49-x^2=0 ==>x=-7,7
area= integral[-7,7]49-x^2- 0 dx
= [-7,7]49x -(1/3)x^3
=(49*7 -(1/3)*7^3 ) -(-49*7 +(1/3)*7^3 )
=2*(49*7 -(1/3)*7^3 )
=(4/3)*7^3
=1372/3
y=49-x^2 ,y=b
49-x^2 =b
==>x=sqrt(49 -b) ,-sqrt(49-b)
area between y=49-x^2 ,y=b
=integral[-sqrt(49-b),sqrt(49-b)] 49-x^2 -b dx =(1/2)*1372/3
=[-sqrt(49-b),sqrt(49-b)] 49x-(1/3)x^3 -bx =686/3
=[49sqrt(49-b)-(1/3)(sqrt(49-b))^3 -bsqrt(49-b)] -[-49sqrt(49-b)+(1/3)(sqrt(49-b))^3 +bsqrt(49-b)] =686/3
=2*[49sqrt(49-b)-(1/3)(sqrt(49-b))^3 -bsqrt(49-b)] =686/3
=[49sqrt(49-b)-(1/3)(sqrt(49-b))^3 -bsqrt(49-b)] =343/3
=[49-(1/3)(49-b) -b] sqrt(49-b)=343/3
=[(2/3)(49-b)] sqrt(49-b)=343/3
=[(2/3)(49-b)3/2]=343/3
=[(49-b)3/2]=343/2
=[(49-b)]=(343/2)2/3
=[(49-b)]=49/(2)2/3
=b=49 -49/(2)2/3
=b=49-30.87
=b=18.132
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