The way you stated it, it would not be a closed region.
I will assume it is also bounded by the y-axis, so we are looking at the right-angled triangle in quadrant II
y = 5x + 25
x = (y-25)/5
x^2 = (y^2 - 50y + 625)/25
= y^2 /25 - 2y + 25
Rotating it about the y-axis
V= π∫ x^2 dy from y = 0 to 25
= π∫( y^2/25 - 2y + 25) dy from 0 to 25
= π[ y^3/75 - y^2 + 25y) form 0 to 25
= π(625/3 - 625 + 625 - 0)
= 625π/3
we can check, since the solid is simply a cone with radius 5 and height of 25
V = (1/3) π r^2 h
= (1/3)π(25)(25) = 625π/3
Find the volume of the solid obtained by rotating the region bounded by
y=5x+25, y=0
about the y-axis.
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