1. The base of a certain solid is the triangle with vertices at (-10,5), (5,5),

Question

1. The base of a certain solid is the triangle with vertices at (-10,5), (5,5), and the origin. Cross-sections perpendicular to the y-axis are squares. What is the volume?

I know that the integral has respect to x, the lower bound is 2, and the upper bound is 6. Can someone help me set up the integral itself?

1. The base of a certain solid is the triangle with vertices at (-10,5), (5,5), and the origin. Cross-sections perpendicular to the y-axis are squares. What is the volume? 2. The volume of the solid obtained by rotating the region enclosed by x=8 can be computed using the method of cylindrical shells via an integral. I know that the integral has respect to x, the lower bound is 2, and the upper bound is 6. Can someone help me set up the integral itself? y=x^2, x=2, x=6, y=0 about the line

Explanation / Answer

1.
From problem statement:
{Base Vertices (0,0) & (-10, 5)} ? {Bounding Base Line} = {y = (-1/2)*x} = {x = (-2)*y}
{Base Vertices (0,0) & (5, 5)} ? {Bounding Base Line} = {y = x} = {x = y}
{Base Vertices (-10, 5) & (5, 5)} ? {Bounding Base Line} = {y = 5}

{Cross-section Area ? y-axis} = {Width}*{Height} =
= {y - (-2)*y}*{y - (-2)*y} = ::: (Square cross-section ? height=width)
= {3y}*{3y} =
= 9*y2

{Differential Volume} = dV = {Cross-section Area}*dy = {9*y2}*dy
Volume = integral of {9*y2}*dy..........limit from 0 to 5
Volume, V = 3y3
apply limits
V = 375

Plz post the second question again becoz in 300 points only 1st part is possible

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