Use calculus to find the volume of the following solid S:
The base of S is an elliptical region with boundary curve 9x^2+4y^2=36. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.
To find the volume of the solid S, we will use calculus and the method of cylindrical shells.
First, let's examine the given elliptical region with the equation 9x^2 + 4y^2 = 36. This equation can be rewritten as x^2/4 + y^2/9 = 1, which represents an ellipse centered at the origin with semi-major axis of length 3 and semi-minor axis of length 2.
To understand the cross-sections of S, let's consider a vertical line segment parallel to the y-axis within the elliptical region. This line segment intersects the ellipse at two points, (x, y1) and (x, y2). The length of this line segment is given by y2 - y1.
Since the cross-sections are isosceles right triangles with the hypotenuse lying on the base, we can deduce that the length of each leg of the triangle is equal to (y2 - y1) / sqrt(2).
Now, we want to find the volume of S by considering an infinitesimally thin vertical strip along the x-axis. Let's consider a small segment of length dx along the x-axis within S.
The height of this segment, corresponding to the y-values, varies from y1 to y2. Therefore, the length of the segment (h) is given by h = y2 - y1.
The width of this segment is dx.
The differential volume element, dV, of the cylindrical shell can be calculated as follows:
dV = (height) × (width) × (length)
= h × dx × (circumference)
The circumference of the cylindrical shell, 2πr, can be approximated as 2πx, since the radius (r) is equal to x due to the cylindrical symmetry of the solid.
Thus, the differential volume element becomes:
dV = h × dx × (2πx)
Now, we integrate dV over the interval where the elliptical region is defined to calculate the total volume of S.
∫dV = ∫(y2 - y1) × dx × (2πx)
To find the limits of integration, we use the equation of the ellipse, x^2/4 + y^2/9 = 1. Solving for y, we get:
y1 = -3√(1 - x^2/4)
y2 = 3√(1 - x^2/4)
Therefore, the limits of integration for x will be from -2 to 2, which corresponds to the semi-major axis of the ellipse.
Now, we can rewrite the integral:
V = ∫(y2 - y1) × dx × (2πx)
= ∫(3√(1 - x^2/4) + 3√(1 - x^2/4)) × (2πx) dx
= 6π ∫√(1 - x^2/4) × x dx
To evaluate this integral, we can use a substitution. Let u = 1 - x^2/4. Then, du = -x/2 dx.
Substituting, we have:
V = 6π ∫√u × (-2u) du
= -12π ∫u^(3/2) du
= -12π (2/5)u^(5/2) + C
Applying the limits of integration, we get:
V = -12π (2/5) [(1 - 2^2/4)^(5/2) - (1 - (-2)^2/4)^(5/2)]
Simplifying further, we have:
V = 12π (2/5) [(-3/4)^(5/2) - (1/4)^(5/2)]
= 12π (2/5) [(3/4)^(5/2) - (1/4)^(5/2)]
Thus, the volume of the solid S is 12π (2/5) [(3/4)^(5/2) - (1/4)^(5/2)].