Problem 54 - Calculus Early Transcendentals

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 \(z=1-y-x^{2}\)

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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: $$V = \int_0^1 \int_0^{1-x^2} (1-y-x^2) \, dy \, dx$$

Step 3: 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 \(\frac{29}{30}\).