find the volume of the solid whose bounded by the circle x^2+y^2=4 and whose cross sections perpendicular to the y-axis are isosceles right triangles with one leg in the base.
Please give explanation and steps
To find the volume of the solid bounded by the circle x^2 + y^2 = 4, we need to break it down into infinitesimally small sections and then integrate over those sections.
Step 1: Visualize the problem
The equation x^2 + y^2 = 4 represents a circle with a radius of 2 centered at the origin. The volume we want to find is the solid enclosed within this circle. Additionally, the cross sections perpendicular to the y-axis are isosceles right triangles with one leg in the base.
Step 2: Define the limits of integration
Since we are integrating perpendicular to the y-axis, we will define y as our variable and integrate with respect to y. The limits of integration for y will be from -2 to 2, which represent the y-values of the circle where it intersects the x-axis.
Step 3: Determine the area of each cross section
Each cross section is an isosceles right triangle, and its height is equal to the length of the base. In this case, since the radius of the circle is 2, the base of each triangle is 2. Thus, the area of each cross section is (1/2) * base * height = (1/2) * 2 * 2 = 2.
Step 4: Set up the integral
To find the volume, we integrate the area of each cross section over the limits of integration. The integral is given by:
V = ∫(from -2 to 2) 2 dy
Step 5: Evaluate the integral
Integrating 2 with respect to y over the limits of -2 to 2 gives:
V = 2y |_(from -2 to 2)
= 2(2) - 2(-2)
= 4 + 4
= 8
Step 6: Finalize the answer
The volume of the solid bounded by the given circle and whose cross sections perpendicular to the y-axis are isosceles right triangles with one leg in the base is 8 cubic units.