Answer:
To find the volume bounded by these surfaces using double integrals, you can set up and evaluate the appropriate double integrals over the given regions. Let's go through each part step by step:
(a) Volume bounded by the plane \(x + y + z = 2\) in the first octant:
First, we need to find the region in the xy-plane where \(x + y + z = 2\) intersects the first octant. This region is a triangle with vertices at (0, 0), (2, 0), and (0, 2).
The integral to find the volume is given by:
\[V = \int_0^2 \int_0^{2-x} (2 - x - y) dy dx\]
Now, evaluate this integral.
(b) Volume bounded by the planes \(2x + y + z = 4\), \(x = 0\), \(y = 0\), and \(z = 0\):
The region in the first octant bounded by these planes is a tetrahedron. To find the volume, set up the integral as follows:
\[V = \int_0^2 \int_0^{4-2x} \int_0^{4-2x-y} dz dy dx\]
Evaluate this integral to find the volume.
(c) Volume bounded by the surfaces \(y = x^2\), \(y + z = 4\), and \(z = 0\):
First, find the region in the xy-plane where \(y = x^2\) intersects \(y + z = 4\). This region is a parabolic segment.
The integral for the volume is given by:
\[V = \int_0^2 \int_{x^2}^{4-x} \int_0^{4-x-y} dz dy dx\]
Evaluate this integral to find the volume.
(d) Volume bounded by the cylinder \(y^2 + z^2 = 1\), the plane \(y = x\), and the yz-plane in the first octant:
The region in the xy-plane where the cylinder intersects the first octant is a quarter-circle with radius 1.
The integral for the volume is given by:
\[V = \int_0^1 \int_0^{\sqrt{1-y^2}} \int_0^{x} dz dx dy\]
Evaluate this integral to find the volume.
Please note that calculating these integrals involves some computation, and the final results will provide the volumes of the regions bounded by the given surfaces.
Explanation: