Answer to Question #315088 in Calculus for Nicz

Question #315088

Find the volume of the solid obtained by rotating the region bounded by y = x^2 +1 and y = 9 – x^2 about y = -1.

1
Expert's answer
2022-03-21T09:16:12-0400

Let’s go to new coordinates:

x˜=x\~x=x;

y˜=y+1\~y=y+1.

Now the volume of the solid can be obtained by rotating the region bounded by y˜=x˜2+2\~y=\~x^2+2 and y˜=10x˜2\~y=10-\~x^2 about y˜=0\~y=0.

Interval of integration:

x˜2+2=10x˜2\~x^2+2= 10-\~x^2

x˜=±2\~x=\pm2

V=22π((10x˜2)2(x˜2+2)2)dx˜=V=\int_{-2}^2 \pi((10-\~x^2)^2-(\~x^2+2)^2)d\~x=

22π(9624x˜2)dx˜=1202π(9624x˜2)dx˜=\int_{-2}^2 \pi(96-24\~x^2)d\~x=\frac12 \int_{0}^2 \pi(96-24\~x^2)d\~x=

12π(96x˜8x˜3)02=π(9684)=64π\frac12\pi (96\~x-8\~x^3)|_0^2=\pi(96-8\cdot4)=64\pi

Answer: V=64πV=64\pi .


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