The following curves bound an area: y = 2x + 1, y = -3, and x = 3. Set up the integral to find the VOLUME if this area is revolv
1 answer:
First note the intersections of each pair of lines.
x = 3 ⇒ y = 2•3 + 1 = 7 ⇒ (3, 7)
y = -3 ⇒ -3 = 2x + 1 ⇒ x = -2 ⇒ (-2, -3)
y = -3 and x = 3 ⇒ (3, -3)
Using the disk method, we consider disks with thickness ∆y and radius equal to the horizontal distance between the line y = 2x + 1 (or x = (y - 1)/2) and the axis of revolution, x = 3. Each disk will then contribute a volume of
∆V = π (radius)² (thickness) = π/4 (y - 1)² ∆y
As we let ∆y go to zero and let the number of disks go to infinity, the total volume of the resulting cone will be given by the integral
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Answer:
Step-by-step explanation:
For this case we have this point given (5,-3) and we want to create an slope intercept formula given by:
And this equation needs to be parallel to y= 3x-1 so then the slope needs to be equal m =3 and then we can use the point given to find the intercept b like this:
And the formula for the slope would be given by:
If we plot that point we find ourselves in QIV. The distance along the x axis is 4, and the distance down from that point is -3. If we create a right triangle with that segment, that segment serves as the hypotenuse of the triangle. We need its measure. Using Pythagorean's theorem, 
and 
. We see that c = 5. We need now to find the secant of that right triangle. Secant if the co-identity of cosine which is side adjacent over hypotenuse. That means that secant is the hypotenuse over the side adjacent. So our secant theta = 5/4