Question

Use double integrals to calculate the volume of the following regions. The solid in the first octant bounded by the coordinate planes and the surface \(z=1-y-x^{2}\)

Short answer

The volume of the solid is \(\frac{29}{30}\).

Step-by-step solution

Setting up the double integral

We will set up a double integral using the given equation, \(z = 1 - y - x^2\), over a suitable domain to represent the volume of the solid. We'll first analyze the intersections with the coordinate planes: 1. \(x = 0\): \(z = 1 - y\) 2. \(y = 0\): \(z = 1 - x^2\) 3. \(z = 0\): \(0 = 1 - y - x^2\) Now, we can set up the double integral as follows: $$V = \iint_{D} (1-y-x^2) \, dA$$ where \(D\) is the region determined by the boundary functions.

Determine the limits of integration

To compute the volume, we must determine the limits of integration for \(x\) and \(y\). From the intersection with the \(z=0\) plane, we have \(1 - y - x^2 = 0\). Solving for \(y\), we get \(y = 1 - x^2\). Since \(x\) and \(y\) are non-negative, we have \(0 \leq x \leq 1\). Moreover, the range of \(y\) values is determined by the boundary function \(y = 1 - x^2\), so we have \(0 \leq y \leq 1 - x^2\). Putting the limits of integration together, we have: $$V = \int_0^1 \int_0^{1-x^2} (1-y-x^2) \, dy \, dx$$

Evaluate the double integral

Now, we can evaluate the double integral to compute the volume of the solid: $$V = \int_0^1 \int_0^{1-x^2} (1-y-x^2) \, dy \, dx$$ First, we'll integrate with respect to \(y\): $$V = \int_0^1 \left[ y - \frac{1}{2}y^2 - x^2y \right]_0^{1-x^2} \, dx$$ Now, substitute the limits: $$V = \int_0^1 ((1-x^2) - \frac{1}{2}(1-x^2)^2 - x^2(1-x^2)) \, dx$$ Next, we'll integrate with respect to \(x\): $$V = \left[ x - \frac{1}{3}x^3 - \frac{1}{5}x^5 + \frac{1}{2}x^7 \right]_0^1$$ Finally, substitute the limits: $$V = (1 - \frac{1}{3} - \frac{1}{5} + \frac{1}{2}) - (0) = \frac{29}{30}$$ The volume of the given solid in the first octant is \(\frac{29}{30}\).