Find the volume V of a solid obtained by rotating the region bounded by y = 8x^3. y = 0, x = 1; rotated about x = 2.
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Let's express Volume as the cross section integrated over y.
Let's express cross section as the integral in polar coordinates with radius from 1(distance from x=2 to x=1) to the limit specified by the curve y=8x3y=8x3:
This limit is dependant on y, as we will integrate over y later on, and our maximum radius equals to the maximum distance in x from x = 2 to the curve x=y−1(y)x=y−1(y). Hence, maximum radius equals to 2−y−1(y)2−y−1(y) In polar coordinates dA=rdrdϕdA=rdrdϕ
In y we integrate from y = 0 to the intersection point of the curve y=8x3y=8x3 and x=1:
V=∫80∫2−18y131∫2π0rdϕdrdyV=∫08∫12−18y13∫02πrdϕdrdy
=∫80∫2−18y1312πrdrdy=∫08∫12−18y132πrdrdy
=∫80[πr2]2−18y131dy=∫08[πr2]12−18y13dy
=π∫80[4−0.5y13+164y23−12]dy=π∫08[4−0.5y13+164y23−12]dy
=π[3y−38y43+35∗64y53]80=π[3y−38y43+35∗64y53]08
=π[3∗8−3∗2423+3∗255∗26]=π[3∗8−3∗2423+3∗255∗26]
=π[24−6+0.3]=18.3π=π[24−6+0.3]=18.3π
Let's express cross section as the integral in polar coordinates with radius from 1(distance from x=2 to x=1) to the limit specified by the curve y=8x3y=8x3:
This limit is dependant on y, as we will integrate over y later on, and our maximum radius equals to the maximum distance in x from x = 2 to the curve x=y−1(y)x=y−1(y). Hence, maximum radius equals to 2−y−1(y)2−y−1(y) In polar coordinates dA=rdrdϕdA=rdrdϕ
In y we integrate from y = 0 to the intersection point of the curve y=8x3y=8x3 and x=1:
V=∫80∫2−18y131∫2π0rdϕdrdyV=∫08∫12−18y13∫02πrdϕdrdy
=∫80∫2−18y1312πrdrdy=∫08∫12−18y132πrdrdy
=∫80[πr2]2−18y131dy=∫08[πr2]12−18y13dy
=π∫80[4−0.5y13+164y23−12]dy=π∫08[4−0.5y13+164y23−12]dy
=π[3y−38y43+35∗64y53]80=π[3y−38y43+35∗64y53]08
=π[3∗8−3∗2423+3∗255∗26]=π[3∗8−3∗2423+3∗255∗26]
=π[24−6+0.3]=18.3π=π[24−6+0.3]=18.3π
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